Answer:
θ ∈ {π/4, π/2, 3π/4, 5π/4, 3π/2, 7π/4}
Explanation:
You want the solutions to sin(2θ)sin(θ)=cos(θ) on [0, 2π).
Factors
We can use the identity for sin(2θ) to write the equation in terms of sine and cosine.
(2sin(θ)cos(θ))·sin(θ) = cos(θ) . . . . replace sin(2θ)
2sin(θ)²cos(θ) - cos(θ) = 0 . . . . . . . subtract cos(θ)
cos(θ)(2 sin(θ)² -1) = 0
Zero product rule
The zero product rule tells you a product will be zero only when one or more factors is zero. The zeros of the factors are ...
cos(θ) = 0 ⇒ θ = π/2, 3π/2
2sin(θ)² -1 = 0 ⇒ sin(θ) = ±√(1/2) ⇒ θ = π/4, 3π/4, 5π/4, 7π/4
The solutions are θ ∈ {π/4, π/2, 3π/4, 5π/4, 3π/2, 7π/4}.
__
Additional comment
The solutions are the x-intercepts of the graph of sin(2θ)sin(θ) -cos(θ), the values of x where this expression is zero. A graphing calculator finds them nicely.