Solve Trigonometric Equations

How to solve trigonometric equations? Questions are presented along with detailed solutions and explanations.

Examples with Solutions

Example 1

Find all the solutions of the trigonometric equation √ 3 sec(?) + 2 = 0
Solution:
Using the identity sec(?) = 1 / cos(?), we rewrite the equation in the form
cos(?) = - √3 / 2
Find the reference ?r angle by solving the above equation without the minus sign:
cos(?r) = √3 / 2     where ?r is acute and equal to the reference angle .
whose solution is given by
?r = ?/6
Use the reference angle ?r to determine the solutions ?1 and ?2 in the interval [0 , 2?) of the given equation. The equation cos(?) = - √3 / 2 suggests that cos(?) is negative and that means the terminal side of angle ? solution to the given equation is either in quadrants II or III as shown below using the unit circle.

graphical solution of cos(x) = 1/2


Hence the solutions in the interval [0 , 2?):
?1 = ? - ?r = ? - ?/6 = 5?/6
?2 = ? + ?r = ? + ?/6 = 7?/6

Use the solutions in the interval [0 , 2?) to find all solutions by adding multiples of 2? as follows:
?1 = 5?/6 + 2n? , n = 0, ± 1 , ± 2, ...
?2 = 7?/6 + 2n? , n = 0, ± 1 , ± 2, ...

Below are shown the graphical solutions on the interval [0 , 2?)

graphical solution of cos(x) = 1/2.


Example 2

Solve the trigonometric equation 2 sin(?) = -1
Solution:

Rewrite the above equation in simple form as shown below.
sin(?) = -1/2

Find the reference angle ?r angle by solving equation without the minus sign:
sin(?) = 1/2
? is acute and equal to
?r = ?/6

Use the reference angle ?r to determine the solutions ?1 and ?2 on the interval [0 , 2?) of the given equation. The equation sin(?) = - 1 / 2 suggests that sin(?) is negative and that means the terminal side of angle ? is either in quadrants III or VI as shown in the unit circle below.

graphical solution of sin(x) = - 1/2


Hence the solutions:
?1 = ? + ?r = 7?/6
?2 = 2? - ?r = 11?/6

Use the solutions on the interval [0 , 2?) to find all solutions by adding multiples of 2? as follows:
?1 = 7?/6 + 2n? , n = 0, ± 1 , ± 2, ...
?2 = 11?/6 + 2n? , n = 0, ± 1 , ± 2, ...

graphical solution of cos(x) = 1/2.


Example 3

Solve the trigonometric equation √2 cos(3x + ?/4) = - 1

Solution:
Let ? = 3x + ?/4 and rewrite the equation in simple form.
√2 cos(?) = - 1
cos(?) = -1/√2

Find the reference angle ?r angle by solving cos(?) = 1/√2 for ?r acute.
?r = ?/4

Use the reference angle ?r to determine the solutions ?1 and ?2 on the interval [0 , 2?) of the given equation. The equation cos(?) = - 1/√2 suggests that cos(?) is negative and that means the terminal side of angle ? is either in quadrants II or III. Hence the two solutions of the equation cos(?) = - 1/√2 on the interval [0 , 2?) are given by
?1 = ? - ?r = 3?/4
?2 = ? + ?r = 5?/4

We now write the general solutions by adding multiples of 2? as follows:
?1 = 3?/4 + 2n? , n = 0, ± 1 , ± 2, ...
?2 = 5?/4 + 2n? , n = 0, ± 1 , ± 2, ...

We now substitute ?1 and ?2 by the expression 3x + ?/4
3x + ?/4 = 3?/4 + 2n?
3x + ?/4 = 5?/4 + 2n?

and solve for x to obtain the solutions for x.
x = ?/6 + 2n?/3 , n = 0, ± 1 , ± 2, ...
x = ?/3 + 2n?/3 , n = 0, ± 1 , ± 2, ...


Example 4

Solve the trigonometric equation - 2 sin 2x - cos x = - 1

Solution:
The above equation may be factored if all trigonometric functions included in that equation are the same. So using the identity sin 2x = 1 - cos 2x, we can rewrite the above equation using the same trigonometric function cos x as follows:
- 2 (1 - cos 2x) - cos x = - 1

Simplify and rewrite as
2 cos 2x - cos x - 1 = 0

Factor the left hand side
(2 cos x + 1)(cos x - 1) = 0

Hence the two equations to solve
(1) 2 cos x + 1 = 0 and (2) cos x - 1 = 0

Solve equation (1) using the reference angle as was done in the examples above.
cos x = -1/2
x1 = 2?/3 + 2n? , n = 0, ± 1 , ± 2, ...
x2 = 4?/3 + 2n? , n = 0, ± 1 , ± 2, ...

Solve equation (2)
cos x = 1
x3 = 2n? , n = 0, ± 1 , ± 2, ...


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