- #1
phospho
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Solve for values of θ in the interval 0 ≤ θ ≤ 360,
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?
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?