You can put this solution on YOUR website! What is sin(2θ)sinθ=cosθ using the interval 0 ≤ θ < 2 π?
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sin(2θ)sinθ=cosθ
2sinθcosθsinθ=cosθ
2sinθcosθsinθ-cosθ=0
cosθ(2sin^2θ-1)=0
cosθ=0
θ=π/2 and 3π/2
or
2sin^2θ-1=0
sin^2θ=1/2
sinθ=±1/√2=±√2/2
θ=π/4 and 3π/4 (in quadrants I and II where sin>0)
and 5π/4 and 7π/4 (in quadrants II and IV where sin<0)
..
θ=π/2, 3π/2, π/4, 3π/4, 5π/4 and 7π/4
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