Solve 2cos2(x)+3sin(2x)=22\cos^2(x) + \sqrt{3}\sin(2x) = 22cos2(x)+3sin(2x)=2 for x∈[0,2π]x \in [0, 2\pi]x∈[0,2π].
x=π3,4π3x = \frac{\pi}{3}, \frac{4\pi}{3}x=3π,34π
x=π6,5π6,7π6,11π6x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}x=6π,65π,67π,611π
x=π3,2π3,4π3,5π3x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}x=3π,32π,34π,35π
x=0,π3,5π3,2πx = 0, \frac{\pi}{3}, \frac{5\pi}{3}, 2\pix=0,3π,35π,2π