Solve for θ\thetaθ in [0,2π)[0, 2\pi)[0,2π) given 4cos2(θ)−3=04\cos^2(\theta) - 3 = 04cos2(θ)−3=0.
π6,5π6,7π6,11π6\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}6π,65π,67π,611π
π3,2π3,4π3,5π3\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}3π,32π,34π,35π
π4,3π4,5π4,7π4\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}4π,43π,45π,47π
π6,11π6\frac{\pi}{6}, \frac{11\pi}{6}6π,611π