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

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.

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 5 + 17 2 – = Simplify. 4.56 or 0.44 Use a calculator.

Use the quadratic formula EXAMPLE 5 Use the quadratic formula x = cos –1 4.56 x = cos –1 0.44 Use inverse cosine. No solution 1.12 Use a calculator, if possible. ANSWER In the interval 0 ≤ x ≤ π, the only solution is x 1.12.

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 = 0 or cos x + 1 = 0 Zero product property cos x = 0 or = –1 Solve for cos x. On the interval 0 ≤ x <2π, cos x = 0 has two solutions: x π 2 = or 3π =

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

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.

GUIDED PRACTICE for Examples 4, 5, and 6 Find the general solution of the equation. 4. sin3 x – sin x = 0 5. 1 – cos x = sin x 3 0 + n π or + n π ANSWER π 2 2π 3 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 7. tan2 x – sin x tan2 x = 0 ANSWER π 4 3π , 0, π or ANSWER π 2