Sin^2 x/3+6cos x/3=1

Решите задачу:

$\sin {}^{2} ( \frac{x}{3} ) + 6 \cos( \frac{x}{3} ) = 1 \\ 1 - \cos {}^{2} ( \frac{x}{3} ) + 6 \cos( \frac{x}{3} ) = 1 \\ - \cos {}^{2} ( \frac{x}{3} ) + 6 \cos(x) = 0 \\ \cos( \frac{x}{3} ) ( - \cos( \frac{x}{3} ) + 6) = 0 \\ \cos( \frac{x}{3} ) = 0 \: \: and \: \: \: \: \: - \cos( \frac{x}{ 3} ) + 6 = 0 \\ \frac{x}{3} = \frac{\pi}{2} + \pi \: k \: \: \: \: \: \: and \: \: \: \: \cos( \frac{x}{3} ) = 6\\ x = \frac{3\pi}{2} + 3\pi \: k \: \: \: and \: \: \: x\in \varnothing$