7.4 Solving Trigonometric Equations
Recall when solving basic equations that the equation must be set equal to zero. Same
holds true when solving trig equations. But there is more to solving trig equations. You
have to consider the quadrant in which the value of the trig function lies in. In addition,
you might have to find all possible solutions to trig equations. For instance, let’s
consider the equation sin x = ½ . Notice the value is positive. So where are the values of
the sine function positive? Referring back to Unit Circle or the graph of the sine function
you will see that the sine function is positive in quadrants I and II. Now you have to go
back to the common right triangle ratios to see which angle you can take the sine of in
order to get ½ . You will see that sin 30˚ = ½ . But the 30˚ represents the reference angle
to assist you in determining the angle in standard position. Now you want to draw a right
triangle with the x-axis having a base angle measure of 30˚.
30˚
30˚
Examining QI, the angle in standard position is 30˚. Examining QII, the angle in
standard position can be obtained by subtracting 30 from 180, yielding 150˚.
Finally, x = 30˚ and 150˚, or in radian measure x = π/6 and 5π/6.
If we want to find all possible solutions, you have to know the length of the period of that
particular trig function, which in this case the sine function has a period length of 2π.
Therefore, x = π/6 ± 2π and 5π/6 ± 2π.
2
.
2
The cosine function is positive in QI and QIV. Going back to the common right triangle
ratios you will see that this value can be obtained by taking cos π/4 or cos 45˚. Drawing
this reference angle in QI and QIV with the x-axis, we get
Example Solve for x on [0, 2π]: cos x =
45˚
45˚
Examining QI, the angle in standard position is 45˚. Examining QIV, the angle in
standard position can be obtained by subtracting 45 from 360, yielding 315˚.
Finally, x = 45˚ and 315˚, or in radian measure x = π/4 and 7π/4.
Try the following:
Solve on [0, 2π].
3
1. sin x =
2
2. cos x = - ½
3. tan x = -1
Answers: π/3, 2π/3 ; 2π/3, 4π/3 ; 3π/4, 7π/4
Example Find all solutions in radian measure: tan x =  3
The tangent function is negative in QII and QIV. Going back to the common right
triangle ratios you will see that this value can be obtained by taking tan π/3 or tan 60˚.
Drawing this reference angle in QII and QIV with the x-axis, we get
60˚
60˚
Examining QII, the angle in standard position is 120˚. Examining QIV, the angle in
standard position can be obtained by subtracting 60 from 360, yielding 300˚.
Finally, x = 120˚ and 300˚, or in radian measure x = 2π/3 and 5π/3. But we want to find
all solutions. The period length for tangent is π. Therefore, the answer to tan x =  3 is
x = 2π/3 ± π and 5π/3 ± π, or more particularly 2π/3 ± π.
Try the following:
Find all solutions in radian measure.
2
1. sin x = 
2
2. cos x =
3
2
3. tan x =
3
3
Answers: 5π/4 ± 2π and 7π/4 ± 2π ; π/6 ± 2π and 11π/6 ± 2π ; π/6 ± π
If the angular expression is some multiple of x, then you will have one more step to do at
the end.
Example Find all solutions in radian measure: cos 3x = - ½ .
The cosine function is negative in QII and QIII. Going back to the common right triangle
ratios you will see that this value can be obtained by taking cos π/3 or cos 60˚. Drawing
this reference angle in QII and QIII with the x-axis, we get
60˚
60˚
Examining QII, the angle in standard position is 120˚. Examining QIII, the angle in
standard position can be obtained by adding 60 to 180, yielding 240˚.
Finally, 3x = 120˚ and 240˚, or in radian measure 3x = 2π/3 and 4π/3. But we want to
find all solutions. The period length for cosine is 2π.
Now the answer to cos 3x = - ½ is 3x = 2π/3 ± 2π and 4π/3 ± 2π.
The last step here is to divide by 3 in order to isolate the variable x. Be sure to also
divide 2π by 3. Finally, x = 2π/9 ± 2π/3 and 4π/9 ± 2π/3.
If the problem had asked you to find the solutions on the interval [0, 2π], then you would
have to add the period change to the angle measures until you reach and not exceed 2π.
Try the following:
1. Find all solutions in radian measure: tan 2x =
3.
2. Solve for x on [0, 2π]: sin 4x = - ½ .
Answers: π/6 ± π/2 ; 7π/24 ± π/2 and 11π/24 ± π/2
At times you will need to factor in order to solve for the variable. Be sure not to divide
both sides of the equal sign to cancel a trig function. You could lose answers by
dividing. Just as in factoring equations, you might have to using the quadratic formula to
get the variable x by itself.
Example
Solve for x on [0, 2π]: (sin 2x)(tan 2x) + sin 2x = 0
(sin 2x)(tan 2x) + sin 2x = 0
Factor what is common
(sin 2x)(tan 2x +1) = 0
Set each factor equal to zero and solve for x as you would have done on previous
examples and problems.
sin 2x = 0
tan 2x +1 = 0
2x = 0 ± 2π, π ± 2π
tan 2x = -1
x = 0 ± π, π/2 ± π
2x = 3π/4 ± π
x = 0, π/2, π, 3π/2, 2π
x = 3π/8 ± π/2
x = 3π/8, 7π/8, 11π/8, 15π/8
Example Solve for x on [0, 2π]: 3sin2(2x) – sin(2x) – 1 = 0
3sin2(2x) – sin(2x) – 1 = 0
There are no common factors throughout the equation, and the equation is not factorable.
Instead, we have to use quadratic formula to assist in solving for x.
sin(2x) =
(1)  (1)2  4(3)(1)
2(3)
sin(2x) =
1  1  12
1  13
=
6
6
sin(2x) =
1  13
 0.7676
6
2x = 0.8751 ± 2π , 2.374 ± 2π
x = 0.44 ± π , 1.187± π
x = 0.44 , 1.187 , 3.582 , 4.329
sin(2x) =
1  13
 0.4343
6
2x = 3.5759 ± 2π , 5.8489 ± 2π
x = 1.788 ± π , 2.924 ± π
x = 1.788 , 2.924 , 4.930 , 6.066
If you chose to work this out on your calculator to make sure you get the same answers,
the answers by calculator will be accurate to the nearest tenth.
Try the following:
Solve for x on [0, 2π].
1. sin2(2x) = ¼
2. cos2(2x) – 3cos(2x) – 2 = 0
Answers: 0.26, 1.31, 1.83, 2.88, 3.40, 4.45, 4.97, 6.02 ; 1.08, 2.06, 4.23, 5.20