Answer:
see the first attachment
Explanation:
The equation can be factored as ...
sin(x)·(2·sin(x) -1) = 0
The equation will be true for ...
sin(x) = 0 ⇒ x = 0, π, 2π . . . . x = 2π is included in your domain
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2sin(x) -1 = 0
sin(x) = 1/2 ⇒ x = π/6, 5π/6