Answer:
θ ∈ {π/4, π/2, 3π/4, 5π/4, 3π/2, 7π/4}
add 2nπ for other solutions, where n = any integer
Explanation:
You want the solutions to sin(2θ)sin(θ) = cos(θ) for all θ and on the interval [0, 2π).
Solution
The equation is best expressed in the form f(θ) = 0, where f(θ) contains only sin(θ) and cos(θ). Using the trig identity for sin(2θ), we can write the equation as ...
(2sin(θ)cos(θ))·sin(θ) - cos(θ) = 0
Factoring, we have ...
cos(θ)(2sin²(θ) -1) = 0
The solutions will be the values of θ that make these factors zero:
cos(θ) = 0 ⇒ θ = odd multiples of π/2
sin²(θ) -1 = 0 ⇒ sin(θ) = ±√(1/2) ⇒ θ = odd multiples of π/4
a. 0 to 2π
The values of θ that satisfy the equation on the interval 0 ≤ θ < 2π are ...
θ ∈ {π/4, π/2, 3π/4, 5π/4, 3π/2, 7π/4}
These are the solutions shown in the attachment.
b. All θ
The sine and cosine functions have period 2π, so these solutions will repeat at intervals of 2π:
θ ∈ {π/4, π/2, 3π/4, 5π/4, 3π/2, 7π/4} +2nπ . . . . n is any integer