Solve for θ\thetaθ in [0,2π)[0, 2\pi)[0,2π) given 2sin2θ=12\sin^2\theta = 12sin2θ=1.
π/4,3π/4,5π/4,7π/4\pi/4, 3\pi/4, 5\pi/4, 7\pi/4π/4,3π/4,5π/4,7π/4
π/4,7π/4\pi/4, 7\pi/4π/4,7π/4
π/4,3π/4\pi/4, 3\pi/4π/4,3π/4
π/2,3π/2\pi/2, 3\pi/2π/2,3π/2