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2sinθ = cosecθ

now if I do:

2sinθ = 1/(sinθ) and multiply by sinθ

I can solve sinθ = ±√(1/2)

but if I solve 2sinθ - 1/(sinθ) = 0 and factorise to get sinθ(2 - (1/sin^2θ)) = 0

I get sinθ = 0 which gives me more solutions then needed. I've drawn the graphs of both functions and they meet where the solutions of θ are such that sinθ = ±√(1/2) , then why do I get more solutions if I factorise?