SAT TRIGONOMETRIC EQUATIONS
Example Solve the following equation:
8 sin x – 6 = 2 sin x – 3, 0° ≤ 𝑥 ≤ 360°.
Solution
8 sin x –2 sin x = 6 – 3
6 sin x = 3
1
sin x =
2
1
Related angle: 𝛼 = sin−1 ( ) = 30°.
2
sin x > 0, therefore there are 2 values of x, in Quadrants I and II. Hence,
𝑥 = α = 30° or 𝑥 = 180° − α = 150°.
Example Express, in degrees, the measure of the obtuse angle that
satisfies the equation 2 cot 𝜃sin 𝜃 + 1 = 0.
Solution Replace cot 𝜃 by
2
cos 𝜃
sin 𝜃
cos 𝜃
sin 𝜃
:
∙sin 𝜃 + 1 = 0
2cos 𝜃 + 1 = 0
1
cos 𝜃 = −
2
1
Related angle: 𝛼 = cos −1 ( ) = 60°.
2
cos 𝜃 < 0, therefore there are 2 values of 𝜃, in Quadrants II and III.
Hence, 𝜃 = 180° − α = 120° or 𝜃 = 180° + α = 240°.
The obtuse angle equals 120°.
Example The value of x between 180and 270that satisfies the
equation tan x = cot x is
(A) 200° (B) 210° (C) 225° (D) 240° (E) 250°

Solution: (C)
tan x =
1
tan 𝑥
,
tan2 x = 1
tan x =± 1
Related angle: 𝛼 = tan−1 1 = 45°.
In quadrant III, x = 180° + α = 225.
Example Find all values of x between 0and 180that satisfy the
equation 2 sin2 x + 3 cos x = 0.
Solution Substitute sin2 x = 1 – cos2 x:
2 – 2cos2 x + 3cos x = 0
2cos2 x – 3cos x – 2 = 0
(2 cos 𝑥 + 1)(cos 𝑥 − 2) = 0
2 cos 𝑥 + 1 = 0 or cos 𝑥 − 2 = 0
1
cos 𝑥 = − or cos x = 2
2
Reject cos x = 2 since cos x≤ 1.
1
cos 𝑥 = −
2
1
Related angle: 𝛼 = cos −1 ( ) = 60°.
2
cos 𝑥 < 0, therefore there are 2 values of 𝜃, in Quadrants II and III.
Hence, 𝑥 = 180° − α = 120° or 𝑥 = 180° + α = 240°.
Since x is between 0and 180we get: 𝑥 = 120°.
Example How many values of x between 0° and 360° satisfy the
equation 2 sec2 x + 5 tan x = 0?
Solution Substitute sec 2 x = 1 + tan2 x:
2(1 + tan2 x) + 5 tan x = 0
2 tan2 x + 5 tan x + 2 = 0
(2 tan 𝑥 + 1)(tan 𝑥 + 2) = 0
tan x = –0.5 or tan x = –2;
For each of these values of tan x, there are 2 values of x, in Quadrants II
and IV. Hence, there are four solutions.
Example Solve for x, 0≤ 𝑥 ≤ 2π.
2 sin2 x + sin x = 1
Solution
2 sin2 x + sin x – 1= 0
(2sin 𝑥 – 1)(sin 𝑥 + 1) = 0
2sin 𝑥 – 1 = 0 or sin 𝑥 + 1 = 0
sin 𝑥 =
1
2
𝜋 5𝜋
𝑥= ,
6
6
or sin 𝑥 = −1
or 𝑥 =
3𝜋
2
.
Example Solve the following equation:
4 sin3 x – sin x = 0, 0° ≤ 𝑥 ≤ 360°.
Solution
sin x (4 sin2 x –1) = 0.
sin x (2sin x – 1) (2sin x + 1) = 0
1
1
2
2
sin x = 0 or sin x = or sin x = − .
1) sin x = 0, x = 0°, 180°, 360°.
1
2) sin x = .
2
1
Related angle: 𝛼 = sin−1 ( ) = 30°.
2
sin x > 0, therefore there are 2 values of x, in Quadrants I and II. Hence,
𝑥 = α = 30° or 𝑥 = 180° − α = 150°.
1
3) sin x = − .
2
1
Related angle: 𝛼 = sin−1 ( ) = 30°.
2
sin x < 0, therefore there are 2 values of x, in Quadrants III and IV.
Hence, 𝑥 = 180° + α = 210° or 𝑥 = 360° − α = 330°.
Example Solve the following equation:
2 tan2 x cos x = tan2 x, 0° ≤ 𝑥 ≤ 360°.
Solution
2 tan2 x cos x – tan2 x = 0
tan2 x (2cos x – 1) = 0
1
tan x = 0 or cos x = .
2
1) tan x = 0, x = 0°, 180°, 360°.
1
2) cos x = .
2
1
Related angle: 𝛼 = cos −1 ( ) = 60°.
2
cos 𝑥 > 0, therefore there are 2 values of 𝜃, in Quadrants I and IV.
Hence, 𝑥 = α = 60° or 𝑥 = 360°– α = 300°.
Example Solve the following equation:
sin2 2x = sin 2x, 0° ≤ 𝑥 ≤ 360°.
Solution
0° ≤ 𝑥 ≤ 360° ⇒0° ≤ 2𝑥 ≤ 720°.
sin2 2x – sin 2x = 0
sin 2x (sin 2x – 1) = 0
sin 2x = 0 or sin 2x = 1
1) sin 2x = 0
2x = 0°, 180°, 360°, 540°, 720°
x = 0°, 90°,180°, 270°, 360°
2) sin 2x = 1
2x = 0, 90°, 450°,
x = 0°, 45°, 225°.
Example Solve the following equation:
2 cos2 x + 3 cos x + 1 = 0, 0° ≤ 𝑥 ≤ 360°.
Solution
(2 cos x + 1)(cos x + 1) = 0
2 cos x + 1= 0 or
cos x + 1= 0
1
cos x = − or cos x = –1
2
1) cos x = −
1
2
1
Related angle: 𝛼 = cos −1 ( ) = 60°.
2
cos 𝑥 < 0, therefore there are 2 values of 𝜃, in Quadrants II and III.
Hence, 𝑥 = 180° − α = 120° or 𝑥 = 180° + α = 240°.
2) cos x = –1
𝑥 = 180°