Solving more complex trigonometric equations. Trigonometric equations. The Ultimate Guide (2019)
home When solving many mathematical problems , especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. Such problems include, for example, linear and quadratic equations , linear and quadratic inequalities
, fractional equations and equations that reduce to quadratic. The principle of successfully solving each of the mentioned problems is as follows: you need to establish what type of problem you are solving, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps. It is obvious that success or failure in solving a particular problem depends mainly on how correctly the type of equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, it is necessary to have the skills to perform identity transformations
and computing. The situation is different with trigonometric equations.
It is not at all difficult to establish the fact that the equation is trigonometric. Difficulties arise when determining the sequence of actions that would lead to the correct answer. By appearance
equation, it is sometimes difficult to determine its type. And without knowing the type of equation, it is almost impossible to choose the right one from several dozen trigonometric formulas.
To solve a trigonometric equation, you need to try:
1. bring all functions included in the equation to “the same angles”;
2. bring the equation to “identical functions”;
3. factor the left side of the equation, etc. Let's consider basic solution methods.
trigonometric equations
I. Reduction to the simplest trigonometric equations
Solution diagram Step 1. Express trigonometric function
through known components. Step 2.
Find the function argument using the formulas:
cos x = a; x = ±arccos a + 2πn, n ЄZ.
sin x = a; x = (-1) n arcsin a + πn, n Є Z.
tan x = a; x = arctan a + πn, n Є Z.
ctg x = a; x = arcctg a + πn, n Є Z. Step 3.
Find the unknown variable.
Example.
2 cos(3x – π/4) = -√2.
1) Solution.
2) cos(3x – π/4) = -√2/2.
3x – π/4 = ±(π – π/4) + 2πn, n Є Z;
3) 3x – π/4 = ±3π/4 + 2πn, n Є Z.
3x = ±3π/4 + π/4 + 2πn, n Є Z;
x = ±π/4 + π/12 + 2πn/3, n Є Z.
Answer: ±π/4 + π/12 + 2πn/3, n Є Z.
II. Replacing a variable
I. Reduction to the simplest trigonometric equations
Solution diagram Reduce the equation to algebraic form relative to one of the trigonometric functions.
through known components. Denote the resulting function by the variable t (if necessary, introduce restrictions on t).
ctg x = a; x = arcctg a + πn, n Є Z. Write down and solve the resulting algebraic equation.
Step 4. Make a reverse replacement.
Step 5. Solve the simplest trigonometric equation.
Find the unknown variable.
2cos 2 (x/2) – 5sin (x/2) – 5 = 0.
2 cos(3x – π/4) = -√2.
1) 2(1 – sin 2 (x/2)) – 5sin (x/2) – 5 = 0;
2sin 2 (x/2) + 5sin (x/2) + 3 = 0.
2) Let sin (x/2) = t, where |t| ≤ 1.
3) 2t 2 + 5t + 3 = 0;
t = 1 or e = -3/2, does not satisfy the condition |t| ≤ 1.
4) sin(x/2) = 1.
5) x/2 = π/2 + 2πn, n Є Z;
x = π + 4πn, n Є Z.
Answer: x = π + 4πn, n Є Z.
III. Equation order reduction method
I. Reduction to the simplest trigonometric equations
Solution diagram Replace given equation linear, using the formulas for reducing the degree:
sin 2 x = 1/2 · (1 – cos 2x);
cos 2 x = 1/2 · (1 + cos 2x);
tg 2 x = (1 – cos 2x) / (1 + cos 2x).
through known components. Solve the resulting equation using methods I and II.
Find the unknown variable.
cos 2x + cos 2 x = 5/4.
2 cos(3x – π/4) = -√2.
1) cos 2x + 1/2 · (1 + cos 2x) = 5/4.
2) cos 2x + 1/2 + 1/2 · cos 2x = 5/4;
3/2 cos 2x = 3/4;
2x = ±π/3 + 2πn, n Є Z;
x = ±π/6 + πn, n Є Z.
Answer: x = ±π/6 + πn, n Є Z.
IV. Homogeneous equations
I. Reduction to the simplest trigonometric equations
Solution diagram Reduce this equation to the form
a) a sin x + b cos x = 0 (homogeneous equation of the first degree)
or to the view
b) a sin 2 x + b sin x · cos x + c cos 2 x = 0 (homogeneous equation of the second degree).
through known components. Divide both sides of the equation by
a) cos x ≠ 0;
b) cos 2 x ≠ 0;
and get the equation for tan x:
a) a tan x + b = 0;
b) a tan 2 x + b arctan x + c = 0.
ctg x = a; x = arcctg a + πn, n Є Z. Solve the equation using known methods.
Find the unknown variable.
5sin 2 x + 3sin x cos x – 4 = 0.
2 cos(3x – π/4) = -√2.
1) 5sin 2 x + 3sin x · cos x – 4(sin 2 x + cos 2 x) = 0;
5sin 2 x + 3sin x · cos x – 4sin² x – 4cos 2 x = 0;
sin 2 x + 3sin x · cos x – 4cos 2 x = 0/cos 2 x ≠ 0.
2) tg 2 x + 3tg x – 4 = 0.
3) Let tg x = t, then
t 2 + 3t – 4 = 0;
t = 1 or t = -4, which means
tg x = 1 or tg x = -4.
From the first equation x = π/4 + πn, n Є Z; from the second equation x = -arctg 4 + πk, k Є Z.
Answer: x = π/4 + πn, n Є Z; x = -arctg 4 + πk, k Є Z.
V. Method of transforming an equation using trigonometric formulas
I. Reduction to the simplest trigonometric equations
Solution diagram Using all sorts of trigonometric formulas, reduce this equation to an equation solved by methods I, II, III, IV.
through known components. Solve the resulting equation using known methods.
Find the unknown variable.
sin x + sin 2x + sin 3x = 0.
2 cos(3x – π/4) = -√2.
1) (sin x + sin 3x) + sin 2x = 0;
2sin 2x cos x + sin 2x = 0.
2) sin 2x (2cos x + 1) = 0;
sin 2x = 0 or 2cos x + 1 = 0;
From the first equation 2x = π/2 + πn, n Є Z; from the second equation cos x = -1/2.
We have x = π/4 + πn/2, n Є Z; from the second equation x = ±(π – π/3) + 2πk, k Є Z.
As a result, x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.
Answer: x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.
The ability and skill to solve trigonometric equations is very 
important, their development requires significant effort, both on the part of the student and on the part of the teacher.
Many problems of stereometry, physics, etc. are associated with the solution of trigonometric equations. The process of solving such problems embodies many of the knowledge and skills that are acquired by studying the elements of trigonometry.
Trigonometric equations occupy an important place in the process of learning mathematics and personal development in general.
Still have questions? Don't know how to solve trigonometric equations?
To get help from a tutor, register.
The first lesson is free!
website, when copying material in full or in part, a link to the source is required.
Requires knowledge of the basic formulas of trigonometry - the sum of the squares of sine and cosine, the expression of tangent through sine and cosine, and others. For those who have forgotten them or do not know them, we recommend reading the article "".
So, we know the basic trigonometric formulas, it's time to use them in practice. Solving trigonometric equations with the right approach - quite exciting activity, like, for example, solving a Rubik's cube.
Based on the name itself, it is clear that a trigonometric equation is an equation in which the unknown is under the sign of the trigonometric function.
There are so-called simplest trigonometric equations. Here's what they look like: sinx = a, cos x = a, tan x = a. Let's consider how to solve such trigonometric equations, for clarity we will use the already familiar trigonometric circle.
sinx = a
cos x = a
tan x = a
cot x = a
Any trigonometric equation is solved in two stages: we reduce the equation to its simplest form and then solve it as a simple trigonometric equation.
There are 7 main methods by which trigonometric equations are solved.
Variable substitution and substitution method
Solving trigonometric equations through factorization
Reduction to a homogeneous equation
Solving equations through the transition to a half angle
Introduction of auxiliary angle
Solve the equation 2cos 2 (x + /6) – 3sin( /3 – x) +1 = 0
Using the reduction formulas we get:
2cos 2 (x + /6) – 3cos(x + /6) +1 = 0
Replace cos(x + /6) with y to simplify and get the usual quadratic equation:
2y 2 – 3y + 1 + 0
The roots of which are y 1 = 1, y 2 = 1/2
Now let's go in reverse order
We substitute the found values of y and get two answer options:
How to solve the equation sin x + cos x = 1?
Let's move everything to the left so that 0 remains on the right:
sin x + cos x – 1 = 0
Let us use the identities discussed above to simplify the equation:
sin x - 2 sin 2 (x/2) = 0
Let's factorize:
2sin(x/2) * cos(x/2) - 2 sin 2 (x/2) = 0
2sin(x/2) * = 0
We get two equations
An equation is homogeneous with respect to sine and cosine if all its terms are relative to the sine and cosine of the same degree of the same angle. To solve a homogeneous equation, proceed as follows:
a) transfer all its members to the left side;
b) take all common factors out of brackets;
c) equate all factors and brackets to 0;
d) a homogeneous equation of a lower degree is obtained in brackets, which in turn is divided into a sine or cosine of a higher degree;
e) solve the resulting equation for tg.
Solve the equation 3sin 2 x + 4 sin x cos x + 5 cos 2 x = 2
Let's use the formula sin 2 x + cos 2 x = 1 and get rid of the open two on the right:
3sin 2 x + 4 sin x cos x + 5 cos x = 2sin 2 x + 2cos 2 x
sin 2 x + 4 sin x cos x + 3 cos 2 x = 0
Divide by cos x:
tg 2 x + 4 tg x + 3 = 0
Replace tan x with y and get a quadratic equation:
y 2 + 4y +3 = 0, whose roots are y 1 =1, y 2 = 3
From here we find two solutions to the original equation:
x 2 = arctan 3 + k
Solve the equation 3sin x – 5cos x = 7
Let's move on to x/2:
6sin(x/2) * cos(x/2) – 5cos 2 (x/2) + 5sin 2 (x/2) = 7sin 2 (x/2) + 7cos 2 (x/2)
Let's move everything to the left:
2sin 2 (x/2) – 6sin(x/2) * cos(x/2) + 12cos 2 (x/2) = 0
Divide by cos(x/2):
tg 2 (x/2) – 3tg(x/2) + 6 = 0
For consideration, let’s take an equation of the form: a sin x + b cos x = c,
where a, b, c are some arbitrary coefficients, and x is an unknown.
Let's divide both sides of the equation by:
Now the coefficients of the equation, according to trigonometric formulas, have properties sin and cos, namely: their modulus is not more than 1 and the sum of squares = 1. Let us denote them respectively as cos and sin, where - this is the so-called auxiliary angle. Then the equation will take the form:
cos * sin x + sin * cos x = C
or sin(x + ) = C
The solution to this simplest trigonometric equation is
x = (-1) k * arcsin C - + k, where
It should be noted that the notations cos and sin are interchangeable.
Solve the equation sin 3x – cos 3x = 1
The coefficients in this equation are:
a = , b = -1, so divide both sides by = 2
Trigonometric equations are not an easy topic. They are too diverse.) For example, these:
sin 2 x + cos3x = ctg5x
sin(5x+π /4) = cot(2x-π /3)
sinx + cos2x + tg3x = ctg4x
Etc...
But these (and all other) trigonometric monsters have two common and obligatory features. First - you won’t believe it - there are trigonometric functions in the equations.) Second: all expressions with x are found within these same functions. And only there! If X appears somewhere outside, For example, sin2x + 3x = 3, this will already be an equation of mixed type. Such equations require individual approach. We will not consider them here.
We will not solve evil equations in this lesson either.) Here we will deal with the simplest trigonometric equations. Why? Yes because the solution any trigonometric equations consists of two stages. At the first stage, the evil equation is reduced to a simple one through a variety of transformations. On the second, this simplest equation is solved. No other way.
So, if you have problems at the second stage, the first stage does not make much sense.)
What do elementary trigonometric equations look like?
sinx = a
cosx = a
tgx = a
ctgx = a
Here A stands for any number. Any.
By the way, inside a function there may not be a pure X, but some kind of expression, like:
cos(3x+π /3) = 1/2
etc. This complicates life, but does not affect the method of solving a trigonometric equation.
How to solve trigonometric equations?
Trigonometric equations can be solved in two ways. The first way: using logic and the trigonometric circle. We will look at this path here. The second way - using memory and formulas - will be discussed in the next lesson.
The first way is clear, reliable, and difficult to forget.) It is good for solving trigonometric equations, inequalities, and all sorts of tricky non-standard examples. Logic is stronger than memory!)
Solving equations using a trigonometric circle.
We include elementary logic and the ability to use the trigonometric circle. Don't you know how? However... You will have a hard time in trigonometry...) But it doesn’t matter. Take a look at the lessons "Trigonometric circle...... What is it?" and "Measuring angles on a trigonometric circle." Everything is simple there. Unlike textbooks...)
Oh, you know!? And even mastered “Practical work with the trigonometric circle”!? Congratulations. This topic will be close and understandable to you.) What is especially pleasing is that trigonometric circle it doesn't matter what equation you solve. Sine, cosine, tangent, cotangent - everything is the same for him. There is only one solution principle.
So we take any elementary trigonometric equation. At least this:
cosx = 0.5
We need to find X. Speaking in human language, you need find the angle (x) whose cosine is 0.5.
How did we previously use the circle? We drew an angle on it. In degrees or radians. And right away saw trigonometric functions of this angle. Now let's do the opposite. Let's draw a cosine on the circle equal to 0.5 and immediately we'll see corner. All that remains is to write down the answer.) Yes, yes!
Draw a circle and mark the cosine equal to 0.5. On the cosine axis, of course. Like this:
Now let's draw the angle that this cosine gives us. Hover your mouse over the picture (or touch the picture on your tablet), and you'll see this very corner X.
The cosine of which angle is 0.5?
x = π /3
cos 60°= cos( π /3) = 0,5
Some people will chuckle skeptically, yes... Like, was it worth making a circle when everything is already clear... You can, of course, chuckle...) But the fact is that this is an erroneous answer. Or rather, insufficient. Circle connoisseurs understand that there are a whole bunch of other angles here that also give a cosine of 0.5.
If you turn the moving side OA full turn, point A will return to its original position. With the same cosine equal to 0.5. Those. the angle will change by 360° or 2π radians, and cosine - no. The new angle 60° + 360° = 420° will also be a solution to our equation, because
An infinite number of such complete revolutions can be made... And all these new angles will be solutions to our trigonometric equation. And they all need to be written down somehow in response. All. Otherwise, the decision does not count, yes...)
Mathematics can do this simply and elegantly. Write down in one short answer infinite set decisions. Here's what it looks like for our equation:
x = π /3 + 2π n, n ∈ Z
I'll decipher it. Still write meaningfully It’s more pleasant than stupidly drawing some mysterious letters, right?)
π /3 - this is the same corner that we saw on the circle and determined according to the cosine table.
2π is one complete revolution in radians.
n - this is the number of complete ones, i.e. whole rpm It is clear that n can be equal to 0, ±1, ±2, ±3.... and so on. As indicated by the short entry:
n ∈ Z
n belongs ( ∈ ) set of integers ( Z ). By the way, instead of the letter n letters may well be used k, m, t etc.
This notation means you can take any integer n . At least -3, at least 0, at least +55. Whatever you want. If you substitute this number into the answer, you will get a specific angle, which will definitely be the solution to our harsh equation.)
Or, in other words, x = π /3 is the only root of an infinite set. To get all the other roots, it is enough to add any number of full revolutions to π /3 ( n ) in radians. Those. 2πn radian.
All? No. I deliberately prolong the pleasure. To remember better.) We received only part of the answers to our equation. I will write this first part of the solution like this:
x 1 = π /3 + 2π n, n ∈ Z
x 1 - not just one root, but a whole series of roots, written down in a short form.
But there are also angles that also give a cosine of 0.5!
Let's return to our picture from which we wrote down the answer. Here she is:
Hover your mouse over the image and we see another angle that also gives a cosine of 0.5. What do you think it is equal to? The triangles are the same... Yes! He equal to angle X , only delayed in the negative direction. This is the corner -X. But we have already calculated x. π /3 or 60°. Therefore, we can safely write:
x 2 = - π /3
Well, of course, we add all the angles that are obtained through full revolutions:
x 2 = - π /3 + 2π n, n ∈ Z
That's all now.) On the trigonometric circle we saw(who understands, of course)) All angles that give a cosine of 0.5. And wrote down these angles in short mathematical form. The answer resulted in two infinite series of roots:
x 1 = π /3 + 2π n, n ∈ Z
x 2 = - π /3 + 2π n, n ∈ Z
This is the correct answer.
Hope, general principle for solving trigonometric equations using a circle is clear. We mark on the circle the cosine (sine, tangent, cotangent) from given equation, draw the corresponding angles and write down the answer. Of course, we need to figure out what corners we are saw on the circle. Sometimes it's not so obvious. Well, I said that logic is required here.)
For example, let's look at another trigonometric equation:
Please take into account that the number 0.5 is not the only possible number in equations!) It’s just more convenient for me to write it than roots and fractions.
We work according to the general principle. We draw a circle, mark (on the sine axis, of course!) 0.5. We draw all the angles corresponding to this sine at once. We get this picture:
Let's deal with the angle first X in the first quarter. We recall the table of sines and determine the value of this angle. It's a simple matter:
x = π /6
We remember about full turns and, with a clear conscience, write down the first series of answers:
x 1 = π /6 + 2π n, n ∈ Z
Half the job is done. But now we need to determine second corner... It's trickier than using cosines, yes... But logic will save us! How to determine the second angle through x? Yes Easy! The triangles in the picture are the same, and the red corner X equal to angle X . Only it is counted from the angle π in the negative direction. That’s why it’s red.) And for the answer we need an angle, measured correctly, from the positive semi-axis OX, i.e. from an angle of 0 degrees.
We hover the cursor over the drawing and see everything. I removed the first corner so as not to complicate the picture. The angle we are interested in (drawn in green) will be equal to:
π - x
X we know this π /6 . Therefore, the second angle will be:
π - π /6 = 5π /6
Again we remember about adding full revolutions and write down the second series of answers:
x 2 = 5π /6 + 2π n, n ∈ Z
That's all. A complete answer consists of two series of roots:
x 1 = π /6 + 2π n, n ∈ Z
x 2 = 5π /6 + 2π n, n ∈ Z
Tangent and cotangent equations can be easily solved using the same general principle for solving trigonometric equations. If, of course, you know how to draw tangent and cotangent on a trigonometric circle.
In the examples above, I used the table value of sine and cosine: 0.5. Those. one of those meanings that the student knows must. Now let's expand our capabilities to all other values. Decide, so decide!)
So, let's say we need to solve this trigonometric equation:
There is no such cosine value in the short tables. We coldly ignore this terrible fact. Draw a circle, mark 2/3 on the cosine axis and draw the corresponding angles. We get this picture.
Let's look, first, at the angle in the first quarter. If only we knew what x is equal to, we would immediately write down the answer! We don’t know... Failure!? Calm! Mathematics does not leave its own people in trouble! She came up with arc cosines for this case. Do not know? In vain. Find out, It's a lot easier than you think. There is not a single tricky spell about “inverse trigonometric functions” on this link... This is superfluous in this topic.
If you are in the know, just say to yourself: “X is an angle whose cosine is equal to 2/3.” And immediately, purely by the definition of arc cosine, we can write:
We remember about the additional revolutions and calmly write down the first series of roots of our trigonometric equation:
x 1 = arccos 2/3 + 2π n, n ∈ Z
The second series of roots for the second angle is almost automatically written down. Everything is the same, only X (arccos 2/3) will be with a minus:
x 2 = - arccos 2/3 + 2π n, n ∈ Z
And that's it! This is the correct answer. Even easier than with table values. There is no need to remember anything.) By the way, the most attentive will notice that this picture shows the solution through the arc cosine in essence, no different from the picture for the equation cosx = 0.5.
Exactly! General principle That's why it's common! I deliberately drew two almost identical pictures. The circle shows us the angle X by its cosine. Whether it is a tabular cosine or not is unknown to everyone. What kind of angle this is, π /3, or what arc cosine is - that’s up to us to decide.
Same song with sine. For example:
Draw a circle again, mark the sine equal to 1/3, draw the angles. This is the picture we get:
And again the picture is almost the same as for the equation sinx = 0.5. Again we start from the corner in the first quarter. What is X equal to if its sine is 1/3? No problem!
Now the first pack of roots is ready:
x 1 = arcsin 1/3 + 2π n, n ∈ Z
Let's deal with the second angle. In the example with a table value of 0.5, it was equal to:
π - x
It will be exactly the same here too! Only x is different, arcsin 1/3. So what!? You can safely write down the second pack of roots:
x 2 = π - arcsin 1/3 + 2π n, n ∈ Z
This is a completely correct answer. Although it doesn't look very familiar. But it’s clear, I hope.)
This is how trigonometric equations are solved using a circle. This path is clear and understandable. It is he who saves in trigonometric equations with the selection of roots on a given interval, in trigonometric inequalities - they are generally solved almost always in a circle. In short, in any tasks that are a little more difficult than standard ones.
Let's apply knowledge in practice?)
Solve trigonometric equations:
First, simpler, straight from this lesson.
Now it's more complicated.
Hint: here you will have to think about the circle. Personally.)
And now they are outwardly simple... They are also called special cases.
sinx = 0
sinx = 1
cosx = 0
cosx = -1
Hint: here you need to figure out in a circle where there are two series of answers and where there is one... And how to write one instead of two series of answers. Yes, so that not a single root from an infinite number is lost!)
Well, very simple):
sinx = 0,3
cosx = π
tgx = 1,2
ctgx = 3,7
Hint: here you need to know what arcsine and arccosine are? What is arctangent, arccotangent? The most simple definitions. But you don’t need to remember any table values!)
The answers are, of course, a mess):
x 1= arcsin0,3 + 2π n, n ∈ Z
x 2= π - arcsin0.3 + 2
Not everything works out? Happens. Read the lesson again. Only thoughtfully(there is such an outdated word...) And follow the links. The main links are about the circle. Without it, trigonometry is like crossing the road blindfolded. Sometimes it works.)
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
Solving simple trigonometric equations.
Solving trigonometric equations of any level of complexity ultimately comes down to solving the simplest trigonometric equations. And in this best helper again it turns out to be a trigonometric circle.
Let's recall the definitions of cosine and sine.
The cosine of an angle is the abscissa (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.
The sine of an angle is the ordinate (that is, the coordinate along the axis) of a point on the unit circle corresponding to a rotation through a given angle.
The positive direction of movement on the trigonometric circle is counterclockwise. A rotation of 0 degrees (or 0 radians) corresponds to a point with coordinates (1;0)
We use these definitions to solve simple trigonometric equations.
1. Solve the equation
This equation is satisfied by all values of the rotation angle that correspond to points on the circle whose ordinate is equal to .
Let's mark a point with ordinate on the ordinate axis:
Draw a horizontal line parallel to the x-axis until it intersects with the circle. We get two points lying on the circle and having an ordinate. These points correspond to rotation angles in and radians:
If we, leaving the point corresponding to the angle of rotation per radian, go around a full circle, then we will arrive at a point corresponding to the angle of rotation per radian and having the same ordinate. That is, this rotation angle also satisfies our equation. We can make as many “idle” revolutions as we like, returning to the same point, and all these angle values will satisfy our equation. The number of “idle” revolutions will be denoted by the letter (or). Since we can make these revolutions in both positive and negative directions, (or) can take on any integer values.
That is, the first series of solutions to the original equation has the form:
, , - set of integers (1)
Similarly, the second series of solutions has the form:
, Where , . (2)
As you might have guessed, this series of solutions is based on the point on the circle corresponding to the angle of rotation by .
These two series of solutions can be combined into one entry:
If we take (that is, even) in this entry, then we will get the first series of solutions.
If we take (that is, odd) in this entry, then we get the second series of solutions.
2. Now let's solve the equation
Since this is the abscissa of a point on the unit circle obtained by rotating through an angle, we mark the point with the abscissa on the axis:
Draw a vertical line parallel to the axis until it intersects with the circle. We will get two points lying on the circle and having an abscissa. These points correspond to rotation angles in and radians. Recall that when moving clockwise we get a negative rotation angle:
Let us write down two series of solutions:
,
,
(We get to the desired point by going from the main full circle, that is.
Let's combine these two series into one entry:
3. Solve the equation
The tangent line passes through the point with coordinates (1,0) of the unit circle parallel to the OY axis
Let's mark a point on it with an ordinate equal to 1 (we are looking for the tangent of which angles is equal to 1):
Let's connect this point to the origin of coordinates with a straight line and mark the points of intersection of the line with the unit circle. The intersection points of the straight line and the circle correspond to the angles of rotation on and :
Since the points corresponding to the rotation angles that satisfy our equation lie at a distance of radians from each other, we can write the solution this way:
4. Solve the equation
The line of cotangents passes through the point with the coordinates of the unit circle parallel to the axis.
Let's mark a point with abscissa -1 on the line of cotangents:
Let's connect this point to the origin of the straight line and continue it until it intersects with the circle. This straight line will intersect the circle at points corresponding to the angles of rotation in and radians:
Since these points are separated from each other by a distance equal to , we can write the general solution of this equation as follows:
In the given examples illustrating the solution of the simplest trigonometric equations, tabular values of trigonometric functions were used.
However, if the right side of the equation contains a non-tabular value, then we substitute the value into the general solution of the equation:
SPECIAL SOLUTIONS:
Let us mark the points on the circle whose ordinate is 0:
Let us mark a single point on the circle whose ordinate is 1:
Let us mark a single point on the circle whose ordinate is equal to -1:
Since it is customary to indicate values closest to zero, we write the solution as follows:
Let us mark the points on the circle whose abscissa is equal to 0:
5.
Let us mark a single point on the circle whose abscissa is equal to 1:
Let us mark a single point on the circle whose abscissa is equal to -1:
And slightly more complex examples:
1.
The sine is equal to one if the argument is equal to
The argument of our sine is equal, so we get:
Divide both sides of the equality by 3:
Answer:
2.
Cosine is zero if the argument of cosine is
The argument of our cosine is equal to , so we get:
Let’s express , to do this we first move to the right with the opposite sign:
Let's simplify the right side:
Divide both sides by -2:
Note that the sign in front of the term does not change, since k can take any integer value.
Answer:
And finally, watch the video lesson “Selecting roots in a trigonometric equation using a trigonometric circle”
This concludes our conversation about solving simple trigonometric equations. Next time we will talk about how to decide.
home mathematical problems, especially those that occur before grade 10, the order of actions performed that will lead to the goal is clearly defined. Such problems include, for example, linear and quadratic equations, linear and quadratic inequalities, fractional equations and equations that reduce to quadratic ones. The principle of successfully solving each of the mentioned problems is as follows: you need to establish what type of problem you are solving, remember the necessary sequence of actions that will lead to the desired result, i.e. answer and follow these steps.
It is obvious that success or failure in solving a particular problem depends mainly on how correctly the type of equation being solved is determined, how correctly the sequence of all stages of its solution is reproduced. Of course, in this case it is necessary to have the skills to perform identical transformations and calculations.
and computing. The situation is different with trigonometric equations.
It is sometimes difficult to determine its type based on the appearance of an equation. And without knowing the type of equation, it is almost impossible to choose the right one from several dozen trigonometric formulas.
equation, it is sometimes difficult to determine its type. And without knowing the type of equation, it is almost impossible to choose the right one from several dozen trigonometric formulas.
To solve a trigonometric equation, you need to try:
1. bring all functions included in the equation to “the same angles”;
2. bring the equation to “identical functions”;
3. factor the left side of the equation, etc. basic methods for solving trigonometric equations.
trigonometric equations
I. Reduction to the simplest trigonometric equations
Solution diagram Express a trigonometric function in terms of known components.
through known components. Step 2.
Find the function argument using the formulas:
cos x = a; x = ±arccos a + 2πn, n ЄZ.
sin x = a; x = (-1) n arcsin a + πn, n Є Z.
tan x = a; x = arctan a + πn, n Є Z.
ctg x = a; x = arcctg a + πn, n Є Z. Step 3.
Find the unknown variable.
Example.
2 cos(3x – π/4) = -√2.
1) Solution.
2) cos(3x – π/4) = -√2/2.
3x – π/4 = ±(π – π/4) + 2πn, n Є Z;
3) 3x – π/4 = ±3π/4 + 2πn, n Є Z.
3x = ±3π/4 + π/4 + 2πn, n Є Z;
x = ±π/4 + π/12 + 2πn/3, n Є Z.
Answer: ±π/4 + π/12 + 2πn/3, n Є Z.
II. Replacing a variable
I. Reduction to the simplest trigonometric equations
Solution diagram Reduce the equation to algebraic form with respect to one of the trigonometric functions.
through known components. Denote the resulting function by the variable t (if necessary, introduce restrictions on t).
ctg x = a; x = arcctg a + πn, n Є Z. Write down and solve the resulting algebraic equation.
Step 4. Make a reverse replacement.
Step 5. Solve the simplest trigonometric equation.
Find the unknown variable.
2cos 2 (x/2) – 5sin (x/2) – 5 = 0.
2 cos(3x – π/4) = -√2.
1) 2(1 – sin 2 (x/2)) – 5sin (x/2) – 5 = 0;
2sin 2 (x/2) + 5sin (x/2) + 3 = 0.
2) Let sin (x/2) = t, where |t| ≤ 1.
3) 2t 2 + 5t + 3 = 0;
t = 1 or e = -3/2, does not satisfy the condition |t| ≤ 1.
4) sin(x/2) = 1.
5) x/2 = π/2 + 2πn, n Є Z;
x = π + 4πn, n Є Z.
Answer: x = π + 4πn, n Є Z.
III. Equation order reduction method
I. Reduction to the simplest trigonometric equations
Solution diagram Replace this equation with a linear one, using the formula for reducing the degree:
sin 2 x = 1/2 · (1 – cos 2x);
cos 2 x = 1/2 · (1 + cos 2x);
tg 2 x = (1 – cos 2x) / (1 + cos 2x).
through known components. Solve the resulting equation using methods I and II.
Find the unknown variable.
cos 2x + cos 2 x = 5/4.
2 cos(3x – π/4) = -√2.
1) cos 2x + 1/2 · (1 + cos 2x) = 5/4.
2) cos 2x + 1/2 + 1/2 · cos 2x = 5/4;
3/2 cos 2x = 3/4;
2x = ±π/3 + 2πn, n Є Z;
x = ±π/6 + πn, n Є Z.
Answer: x = ±π/6 + πn, n Є Z.
IV. Homogeneous equations
I. Reduction to the simplest trigonometric equations
Solution diagram Reduce this equation to the form
a) a sin x + b cos x = 0 (homogeneous equation of the first degree)
or to the view
b) a sin 2 x + b sin x · cos x + c cos 2 x = 0 (homogeneous equation of the second degree).
through known components. Divide both sides of the equation by
a) cos x ≠ 0;
b) cos 2 x ≠ 0;
and get the equation for tan x:
a) a tan x + b = 0;
b) a tan 2 x + b arctan x + c = 0.
ctg x = a; x = arcctg a + πn, n Є Z. Solve the equation using known methods.
Find the unknown variable.
5sin 2 x + 3sin x cos x – 4 = 0.
2 cos(3x – π/4) = -√2.
1) 5sin 2 x + 3sin x · cos x – 4(sin 2 x + cos 2 x) = 0;
5sin 2 x + 3sin x · cos x – 4sin² x – 4cos 2 x = 0;
sin 2 x + 3sin x · cos x – 4cos 2 x = 0/cos 2 x ≠ 0.
2) tg 2 x + 3tg x – 4 = 0.
3) Let tg x = t, then
t 2 + 3t – 4 = 0;
t = 1 or t = -4, which means
tg x = 1 or tg x = -4.
From the first equation x = π/4 + πn, n Є Z; from the second equation x = -arctg 4 + πk, k Є Z.
Answer: x = π/4 + πn, n Є Z; x = -arctg 4 + πk, k Є Z.
V. Method of transforming an equation using trigonometric formulas
I. Reduction to the simplest trigonometric equations
Solution diagram Using all possible trigonometric formulas, reduce this equation to an equation solved by methods I, II, III, IV.
through known components. Solve the resulting equation using known methods.
Find the unknown variable.
sin x + sin 2x + sin 3x = 0.
2 cos(3x – π/4) = -√2.
1) (sin x + sin 3x) + sin 2x = 0;
2sin 2x cos x + sin 2x = 0.
2) sin 2x (2cos x + 1) = 0;
sin 2x = 0 or 2cos x + 1 = 0;
From the first equation 2x = π/2 + πn, n Є Z; from the second equation cos x = -1/2.
We have x = π/4 + πn/2, n Є Z; from the second equation x = ±(π – π/3) + 2πk, k Є Z.
As a result, x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.
Answer: x = π/4 + πn/2, n Є Z; x = ±2π/3 + 2πk, k Є Z.
The ability and skill to solve trigonometric equations is very important, their development requires significant effort, both on the part of the student and on the part of the teacher.
Many problems of stereometry, physics, etc. are associated with the solution of trigonometric equations. The process of solving such problems embodies many of the knowledge and skills that are acquired by studying the elements of trigonometry.
Trigonometric equations occupy an important place in the process of learning mathematics and personal development in general.
Still have questions? Don't know how to solve trigonometric equations?
To get help from a tutor -.
The first lesson is free!
blog.site, when copying material in full or in part, a link to the original source is required.