Solve for θ\thetaθ: 2sin2θ−1+1sinθ+1=0\frac{2}{\sin^2 \theta - 1} + \frac{1}{\sin \theta + 1} = 0sin2θ−12+sinθ+11=0.
sinθ=0\sin \theta = 0sinθ=0
sinθ=1\sin \theta = 1sinθ=1
No solution
sinθ=−1\sin \theta = -1sinθ=−1