How do you solve cos(π/2 + 2x) = √2 sin x and find its roots on [−3π, −3π/2]?

Question

a) Solve cos(π/2 + 2x) = √2 sin x. b) Find the roots in the closed interval [−3π, −3π/2].

Answer choices

1. a) 2πk, ±π/4 + 2πk, k ∈ ℤ; b) −2π, −11π/4, −13π/4.
2. a) πk, ±π/4 + 2πk, k ∈ ℤ; b) −2π, −9π/4, −7π/4, −3π.
3. a) πk, ±3π/4 + πk, k ∈ ℤ; b) −2π, −9π/4, −11π/4, −3π.
4. a) πk, ±3π/4 + 2πk, k ∈ ℤ; b) −2π, −11π/4, −3π.

Explanation

Using cos(π/2 + 2x) = −sin 2x, we obtain:

−2 sin x cos x = √2 sin x,

sin x (−2 cos x − √2) = 0.

There are two cases:

1. sin x = 0, giving x = πk;
2. cos x = −√2/2, giving x = ±3π/4 + 2πk.

The first family contributes −3π and −2π to [−3π, −3π/2]. The second contributes −11π/4. The left endpoint −3π is included because the interval is closed.