2sin2θ=1+cosθ ⟹ 2(1−cos2θ)=1+cosθ2\sin^{2}\theta = 1 + \cos\theta \implies 2(1 - \cos^{2}\theta) = 1 + \cos\theta2sin2θ=1+cosθ⟹2(1−cos2θ)=1+cosθ
2−2cos2θ=1+cosθ2 - 2\cos^{2}\theta = 1 + \cos\theta2−2cos2θ=1+cosθ
2−2cos2θ−1−cosθ=02 - 2\cos^{2}\theta - 1 - \cos\theta = 02−2cos2θ−1−cosθ=0
2cos2θ+cosθ−1=02\cos^{2}\theta + \cos\theta - 1 = 02cos2θ+cosθ−1=0
2cos2θ+2cosθ−cosθ−1=0 ⟹ 2cosθ(cosθ+1)−1(cosθ+1)=02\cos^{2}\theta + 2\cos\theta - \cos\theta - 1 = 0 \implies 2\cos\theta(\cos\theta + 1) - 1(\cos\theta + 1) = 02cos2θ+2cosθ−cosθ−1=0⟹2cosθ(cosθ+1)−1(cosθ+1)=0
(2cosθ−1)(cosθ+1)=0 ⟹ cosθ=12(2\cos\theta - 1)(\cos\theta + 1) = 0 \implies \cos\theta = \frac{1}{2}(2cosθ−1)(cosθ+1)=0⟹cosθ=21
θ=cos−112=60∘\theta = \cos^{-1} \frac{1}{2} = 60^{\circ}θ=cos−121=60∘