1. Solve . Original equation
Solve by Isolating Trigonometric Expressions
Solve
Original equation
Subtract 3cos x from each side to isolate the trigonometric expression.
Solve for cos x.
Example 1

2. Solve by Isolating Trigonometric Expressions
The period of cosine is 2π, so you only need to find solutions on the interval [0, 2π). The solutions on this interval are The solutions on the interval (–∞, ∞) are then found by adding integer multiples 2π. Therefore, the general form of the solutions is x = , where n is an integer.
Answer:
Example 1

3. Solve sin x = – sin x. A. B. C. D.
Example 1

4. 3 tan2 x – 4 = –3 Original equation 3 tan2 x = 1 Add 4 to each side.
Solve by Taking the Square Root of Each Side
Solve 3 tan2 x – 4 = –3.
3 tan2 x – 4 = –3 Original equation
3 tan2 x = 1 Add 4 to each side.
Divide each side by 3.
Take the square root of each side.
Rationalize the denominator.
Example 2

5. Solve by Taking the Square Root of Each Side
The period of tangent is π. On the interval [0, π), tan x = when x = and tan x = when x = The solutions on the interval (–∞, ∞) have the general form , where n is an integer.
Answer:
Example 2

6. Solve 5 tan2x – 15 = 0. A. B. C. D.
Example 2

7. A. Find all solutions of on the interval [0, 2π).
Solve by Factoring
A. Find all solutions of on the interval [0, 2π).
Original equation
Isolate the trigonometric terms.
Factor.
Zero Product Property
Example 3

8. On the interval [0, 2π), the equation has solutions .
Solve by Factoring
Solve for cos x.
Solve for x on [0, 2π).
On the interval [0, 2π), the equation has solutions
Answer:
Example 3

9. Solve by Factoring
B. Find all solutions of 2sin2x + sinx – 1 = 0 on the interval [0, 2π).
Original equation
Factor.
Zero Product Property
Solve for sin x.
Solve for x on [0, 2π).
Example 3

10. Solve by Factoring
On the interval [0, 2π), the equation 2sin2x + sinx – 1 = 0 has solutions
Answer:
Example 3

11. Find all solutions of 2 tan4 x – tan2 x – 15 = 0 on the interval [0, π).B. C. D.
Example 3

12. Trigonometric Functions of Multiple Angles
PROJECTILES A projectile is sent off with an initial speed vo of 350 m/s and clears a fence 3000 m away. The height of the fence is the same height as the initial height of the projectile. If the distance the projectile traveled is given by , find the interval of possible launch angles to clear the fence.
Example 4

13. Definition of inverse sine.
Trigonometric Functions of Multiple Angles
Original formula
d = 3000 and v0 = 350
Simplify.
Multiply each side by 9.8.
Divide each side by 122,500.
Definition of inverse sine.
Example 4

14. sin–10.24 = 2 Definition of inverse sine
Trigonometric Functions of Multiple Angles
Recall from Lesson 4-6 that the range of the inverse sine function is restricted to acute angles of θ in the interval [–90°, 90°]. Since we are finding the inverse sine of 2θ instead of θ, we need to consider angles in the interval [–2(90°), 2(90°)] or [–180°, 180°]. Use your calculator to find the acute angle and the reference angle relationship sin (180° − θ) = sin θ to find the obtuse angle.
sin–10.24 = 2 Definition of inverse sine
13.9° or 166.1° = 2 sin–1(0.24) ≈13.9° and sin(180° – 13.9°) = 166.1°
7.0° or 83.1° =  Divide by 2.
Example 4

15. Trigonometric Functions of Multiple Angles
The interval is [7.0°, 83.1°]. The ball will clear the fence if the angle is between 7.0° and 83.1°.
Answer: 7.0° ≤  ≤ 83.1°
CHECK Substitute the angle measures into the original equation to confirm the solution.
Original formula
 = 7.0° or  = 83.1° 
Use a calculator.
Example 4

16. GOLF A golf ball is sent off with an initial speed vo of 36 m/s and clears a small barricade 70 m away. The height of the barricade is the same height as the initial height of the ball. If the distance the ball traveled is given by , find the interval of possible launch angles to clear the barricade.
A. 1.6° ≤  ≤ 88.5°
B. 3.1° ≤  ≤ 176.9°
C. 16.0° ≤  ≤ 74.0°
D. 32° ≤  ≤ 148.0°
Example 4

17. sin2 x – sin x + 1 = cos2 x Original equation
Solve by Rewriting Using a Single Trigonometric Function
Find all solutions of sin2 x – sin x + 1 = cos2 x on the interval [0, 2π).
sin2 x – sin x + 1 = cos2 x Original equation
–cos2 x + sin2 x – sin x + 1 = 0 Subtract cos2 x from each side.
–(1 – sin2 x) + sin2 x – sin x + 1 = 0 Pythagorean Identity
2sin2 x – sin x = 0 Simplify.
sin x (2sin x – 1) = 0 Factor.
Example 5

18. sin x = 0 2sin x – 1 = 0 Zero Product Property
Solve by Rewriting Using a Single Trigonometric Function
sin x = 0 2sin x – 1 = 0 Zero Product Property
2sin x = 1 Solve for sin x.
Solve for x on [0, 2π).
x = 0, π
Example 5

19. Solve by Rewriting Using a Single Trigonometric Function
Answer:
CHECK The graphs of Y1 = sin2 x – sin x + 1 and Y2 = cos2 x intersect at on the interval [0, 2π) as shown. 
Example 5

20. Find all solutions of 2sin2x = cosx + 1 on the interval [0, 2).B. C. D.
Example 5

21. Find all solutions of sin x – cos x = 1 on the interval [0, 2π).
Solve by Squaring
Find all solutions of sin x – cos x = 1 on the interval [0, 2π).
sin x – cos x = 1 Original equation
sin x = cos x + 1 Add cos x to each side.
sin2 x = cos2 x + 2cos x + 1 Square each side.
1 – cos2 x = cos2 x + 2cos x + 1 Pythagorean Identity
0 = 2cos2 x + 2cos x Subtract 1 – cos2x from each side.
0 = cos2 x + cos x Divide each side by 2.
0 = cos x(cos x + 1) Factor.
Example 6

22. cos x = 0 cos x + 1 = 0 Zero Product Property
Solve by Squaring
cos x = 0 cos x + 1 = 0 Zero Product Property
cos x = –1 Solve for cos x.
, x = π Solve for x on [0, 2).
Original formula
Substitute
sin π – cos π = 1 
Simplify.
Example 6

23. Therefore, the only valid solutions are on the interval .
Solve by Squaring
Therefore, the only valid solutions are on the interval .
Answer:
Example 6

24. Find all solutions of 1 + cos x = sin x on the interval [0, 2π).B. C. D.
Example 6

25. End of the Lesson