Solve by Isolating Trigonometric Expressions

Solve .

Original equation

Subtract 3cos x from each side to isolate the trigonometric expression.

Solve for cos x.

Solve by Isolating Trigonometric Expressions

Answer:

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.

Solve sin x + = – sin x.

A.

B.

C.

D.

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.

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:

Solve 5 tan2x – 15 = 0.

A.

B.

C.

D.

Solve by Factoring

A. Find all solutions of on the interval [0, 2π).

Original equation

Isolate the trigonometric terms.

Factor.

Zero Product Property

Solve by Factoring

Solve for x on [0, 2π).

On the interval [0, 2π), the equation

has solutions .

Answer:

Solve for cos x.

Solve by Factoring

B. Find all solutions of 2sin2x + sinx – 1 = 0 on the interval [0, 2π).

Original equation

Factor.

Zero Product PropertySolve for sin x.

Solve for x on [0, 2π).

Solve by Factoring

Answer:

On the interval [0, 2π), the equation 2sin2x + sinx – 1 = 0

has solutions .

Find all solutions of 2 tan4 x – tan2 x – 15 = 0 on the interval [0, π).

A.

B.

C.

D.

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.

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.

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°= 2sin–1(0.24) ≈13.9° and sin(180° – 13.9°) = 166.1°

7.0° or 83.1° = Divide by 2.

Answer: 7.0° ≤ ≤ 83.1°

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°.

CHECK Substitute the angle measures into the original equation to confirm the solution.

Original formula

Use a calculator.

= 7.0° or = 83.1°

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°

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.

Solve by Rewriting Using a Single Trigonometric Function

sin x = 02sin x – 1= 0Zero Product Property

2sin x = 1Solve for sin x.

Solve for x on [0, 2π).

x = 0, π

Solve by Rewriting Using a Single Trigonometric Function

CHECK The graphs of Y1 = sin2 x – sin x + 1 and

Y2 = cos2 x intersect at on the interval

[0, 2π) as shown.

Answer:

Find all solutions of 2sin2x = cosx + 1 on the interval [0, 2).

A.

B.

C.

D.

Find all solutions of sin x – cos x = 1 on the interval [0, 2π).

sin x – cos x= 1Original equation

sin x= cos x + 1Add cos x to each side.

sin2 x= cos2 x + 2cos x + 1 Square each side.

1 – cos2 x= cos2 x + 2cos x + 1Pythagorean Identity

0= 2cos2 x + 2cos xSubtract 1 – cos2x from each side.

0= cos2 x + cos x Divide each side by 2.

0= cos x(cos x + 1)Factor.

Solve by Squaring

Solve by Squaring

cos x = 0 cos x + 1= 0 Zero Product Property

cos x= –1 Solve for cos x.

Original formula

Simplify.

, x = πSolve for x on [0, 2).

Substitute sin π – cos π = 1

Solve by Squaring

Therefore, the only valid solutions are on the interval .

Answer:

Find all solutions of 1 + cos x = sin x on the interval [0, 2π).

A.

B.

C.

D.