Answer:
θ = {π/6, 5π/6} +2kπ . . . . for any integer k
Explanation:
Multiplying by the product of the denominators, we can simplify this to ...
sin(θ)² +(1+cos(θ))² = 4sin(θ)(1+cos(θ))
sin(θ)² +1 +2cos(θ) +cos(θ)² = 4sin(θ)+4sin(θ)cos(θ)
2 +2cos(θ) = 2sin(θ)(2 +2cos(θ)) . . . . show similar factors
2(2sin(θ) -1)(1 +cos(θ)) = 0 . . . . subtract left side, complete the factoring
These factors are zero when ...
2sin(θ) -1 = 0 ⇒ sin(θ) = 1/2 ⇒ θ = π/6, 5π/6
1 +cos(θ) = 0 . . . . . extraneous solution; makes equation undefined
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Solutions are periodic with period 2π, so the complete solution set is ...
θ = {π/6, 5π/6} +2kπ . . . . for any integer k