Solve for θ∈[0,2π]\theta \in [0, 2\pi]θ∈[0,2π]: 2sin2θ+3cosθ+1=02\sin^2\theta + \sqrt{3}\cos\theta + 1 = 02sin2θ+3cosθ+1=0.
5π6,7π6\frac{5\pi}{6}, \frac{7\pi}{6}65π,67π
π3,5π3\frac{\pi}{3}, \frac{5\pi}{3}3π,35π
2π3,4π3\frac{2\pi}{3}, \frac{4\pi}{3}32π,34π
5π6,11π6\frac{5\pi}{6}, \frac{11\pi}{6}65π,611π