1. Standardized Test Practice
EXAMPLE 4
Standardized Test Practice
SOLUTION
sin3 x – 4 sin x =
Write original equation.
sin x (sin2 x – 4) =
Factor out sin x.
sin x (sin x + 2)(sin x – 2) =
Factor difference of squares.
Set each factor equal to 0 and solve for x, if possible.
sin x
= 0
sin x + 2
= 0
sin x – 2
= 0 x
= 0 or
= π
sin x
= –2
sin x
= 2

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.

3. Use the quadratic formula
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.
cos2 x – 5 cos x + 2 =
Write original equation. =
–(–5) + (–5)2 – 4(1)(2)
2(1) –
cos x
Quadratic formula 2 – =
Simplify.
4.56 or 0.44
Use a calculator.

4. Use the quadratic formula
EXAMPLE 5
Use the quadratic formula x
= cos –1 4.56 x
= cos –
Use inverse cosine.
No solution
1.12
Use a calculator, if possible.
ANSWER
In the interval 0 ≤ x ≤ π, the only solution is x

5. Solve an equation with an extraneous solution
EXAMPLE 6
Solve an equation with an extraneous solution
Solve 1 + cos x = sin x in the interval 0 ≤ x < 2π.
1 + cos x
sin x =
Write original equation.
(1 + cos x)2
(sin x)2 =
Square both sides.
1 + 2 cos x + cos2 x
sin2 x =
Multiply.
1 + 2 cos x + cos2 x
1– cos2 x =
Pythagorean identity
2 cos2 x + 2 cos x =
Quadratic form
2 cos x (cos x + 1) =
Factor out 2 cos x.
2 cos x
= or
cos x + 1
= 0
Zero product property
cos x
= or
= –1
Solve for cos x.
On the interval 0 ≤ x <2π, cos x = 0 has two solutions: x π 2
= or 3π =

6. 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 π 2
= ,
π, and 3π
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 π 2
= sin ?
1 + cos π ?
= sin π 3π 2
1 + cos ?
= sin
1 + 0 ?
= 1
1 + (–1) ?
= 0
1 + 0 ?
= –1 1 
= 1 
= 0 1
= –1

7. EXAMPLE 6
Solve an equation with an extraneous solution
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 3π 2
Graphs of each side of the original equation confirm the solutions.

8. GUIDED PRACTICE
for Examples 4, 5, and 6
Find the general solution of the equation.
4. sin3 x – sin x = 0
– cos x = sin x 3
0 + n π or n π
ANSWER π 2 2π 3
0 + 2n π or n π
ANSWER

9. GUIDED PRACTICE
for Examples 4, 5, and 6
Solve the equation in the interval 0 ≤ x <π.
sin x = csc x
7. tan2 x – sin x tan2 x = 0
ANSWER π 4 3π ,
0, π or
ANSWER π 2