The solutions are: θ = 90º, 270º, 30º, and 150º
Given,
sin2θ = cosθ
sin2 θ = 2sinθ cosθ
2sinθ cosθ = cosθ
2sinθ cosθ - cosθ = 0
factor
cosθ(2sinθ-1) = 0
cosθ = 0 or 2sinθ - 1 = 0
cosθ = 0 when θ=90º or θ=270º (look on the unit circle)
2sinθ - 1 = 0
2sinθ = 1
sinθ = 1/2
look at the unit circle and see that sinθ=1/2 when θ=30º
sin is positive in Quadrant2 as well as Quadrant 1
180º-30º=150º
θ=150º
answers: θ=90º, 270º, 30º, and 150º
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