EXAMPLE 4 Standardized Test Practice

SOLUTION

Write original equation.sin3 x – 4 sin x 0=Factor out sin x.sin x (sin2 x – 4) 0=

Factor difference of squares.sin x (sin x + 2)(sin x – 2) 0=

Set each factor equal to 0 and solve for x, if possible.

sin x = 0

x = 0 or x = π

sin x + 2 = 0

sin x = –2

sin x – 2 = 0

sin x = 2

EXAMPLE 4 Standardized Test Practice

The only solutions in the interval 0 ≤ x ≤ 2π, are x = 0 and x = π.

The general solution is x = 2nπ or x = π + 2nπ where n is any integer.

ANSWER

The correct answer is D.

EXAMPLE 5 Use the quadratic formula

Solve cos2 x – 5 cos x + 2 = 0 in the interval 0 ≤ x <π.

SOLUTION

Because the equation is in the form au2 + bu + c = 0 where u = cos x, you can use the quadratic formula to solve for cos x.

Write original equation.cos2 x – 5 cos x + 2 = 0

Quadratic formulacos x = –(–5) + (–5)2 – 4(1)(2)

2(1)–

Simplify.5 + 17

2–=

Use a calculator. 4.56 or 0.44

EXAMPLE 5 Use the quadratic formula

Use inverse cosine.x = cos –1 4.56 x = cos –1 0.44

Use a calculator, if possible.No solution 1.12

ANSWER

In the interval 0 ≤ x ≤ π, the only solution is x 1.12.

EXAMPLE 6 Solve an equation with an extraneous solution

Solve 1 + cos x = sin x in the interval 0 ≤ x < 2π.

Write original equation.1 + cos x sin x=

Square both sides.(1 + cos x)2 (sin x)2=

Multiply.1 + 2 cos x + cos2 x sin2 x=

Pythagorean identity1 + 2 cos x + cos2 x 1– cos2 x=

Quadratic form2 cos2 x + 2 cos x 0=

Factor out 2 cos x.2 cos x (cos x + 1) = 0

Zero product property2 cos x = 0 or cos x + 1 = 0

cos x = 0 or cos x = –1 Solve for cos x.

On the interval 0 ≤ x <2π, cos x = 0 has two solutions:

x π2= or x 3π

2=

EXAMPLE 6 Solve an equation with an extraneous solution

On the interval 0 ≤ x <2π, cos x = –1 has one solution:

x = π.

Therefore, 1 + cos x = sin x has three possible solutions:

x π2= , π, and

3π2

CHECK To check the solutions, substitute them into the original equation and simplify.

1 + cos x = sin x 1 + cos x = sin x 1 + cos x = sin x

1 + cos π2 = sinπ

2?

1 + cos π?= sin π

3π21 + cos

3π2

?= sin

1 + 0?= 1 1 + (–1)

?= 0 1 + 0

?= –1

1 = 1 0 = 0 1 = –1

EXAMPLE 6 Solve an equation with an extraneous solution

3π2

ANSWER

The apparent solution x = is extraneous because it does not check in the original equation. The only solutions in the interval 0 ≤ x <2π are x = and x = π.π

2

Graphs of each side of the original equation confirm the solutions.

GUIDED PRACTICE for Examples 4, 5, and 6

Find the general solution of the equation.

4. sin3 x – sin x = 0

0 + n π or + n π

ANSWER

π2

5. 1 – cos x = sin x3

2π3

0 + 2n π or + 2n π

ANSWER

GUIDED PRACTICE for Examples 4, 5, and 6

Solve the equation in the interval 0 ≤ x <π.

6. 2 sin x = csc x

ANSWER

π4

3π4

, 0, π or

ANSWER

π2

7. tan2 x – sin x tan2 x = 0