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10 questions from this quiz
2sin2θ−sinθ−1=02\sin^2 \theta - \sin \theta - 1 = 02sin2θ−sinθ−1=0
sin2θ+cos2θ=1\sin^2 \theta + \cos^2 \theta = 1sin2θ+cos2θ=1
θ=θ0+360∘n\theta = \theta_0 + 360^\circ nθ=θ0+360∘n
θ=θ0+180∘n\theta = \theta_0 + 180^\circ nθ=θ0+180∘n
0∘0^\circ0∘ to 360∘360^\circ360∘
To discard extraneous solutions
(2u−1)(u−1)=0(2u - 1)(u - 1) = 0(2u−1)(u−1)=0
30∘,150∘30^\circ, 150^\circ30∘,150∘
0≤3θ≤5400 \leq 3\theta \leq 5400≤3θ≤540
An angle more complicated than just θ\thetaθ
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