Question

Find all solutions of the equation 2 sin 2x cos x - sin 2x = 0 over the interval -pi/2 < x lessthanorequalto pi/2.

a.x = -pi/6, 0, pi/6, pi/2
b.x = -pi/3, -pi/4, pi/4, pi/3
c.x = -pi/4, -pi/6, pi/6, pi/4 d.x = -pi/3, 0, pi/3, pi/2

Answer 1

To solve the equation 2 sin 2x cos x - sin 2x = 0, we can use trigonometric identities to simplify and factor the equation. The solutions within the given interval are x = -pi/2, 0, pi/2, -pi/3, pi/3.

Step-by-step explanation:

To solve the equation 2 sin 2x cos x - sin 2x = 0, we'll use the trigonometric identity sin 2x = 2 sin x cos x. Substituting this into the equation gives us 2(2 sin x cos x) cos x - 2 sin x cos x = 0. Simplifying further, we have 4 sin x cos^2 x - 2 sin x cos x = 0. Factoring out sin x cos x, we get sin x cos x (4 cos x - 2) = 0.

Setting each factor equal to zero gives us two equations: sin x = 0 and 4 cos x - 2 = 0.

For sin x = 0, the solutions within the interval -pi/2 ≤ x ≤ pi/2 are x = -pi/2, 0, pi/2.

For 4 cos x - 2 = 0, we solve for cos x to get cos x = 1/2. The solutions within the given interval are x = -pi/3, pi/3.