Нужна помощь с решением: cos5x + cos7x = sin2x

[latex]cos5x + cos7x = sin2x \\ 2cos \frac{12x}{2}cos \frac{-2x}{2} = sin2x \\ 2cos6xcosx=2sinxcosx\\ cos6xcosx=sinxcosx \\ cosx(cos6x- sinx) = 0 \\ 1. cosx = 0 \\ x = \frac{ \pi}{2} + 2 \pi n \\ 2 cos6x - sinx = 0\\ sin( \frac{ \pi}{2} - 6x ) - sinx = 0 \\ sin( \frac{ \pi}{2} - 6x ) = sinx \\ 2.1 \\2 \pi n+ \frac{ \pi}{2} - 6x = x \\ 2 \pi n + \frac{ \pi}{2} = 7x \\ x = \frac{2 \pi n}{7} + \frac{ \pi}{14} \\ [/latex] [latex]2.2\\ 2 \pi n+ \frac{ 3\pi}{2}+ 6x = x \\ 2 \pi n+ \frac{3 \pi}{2} = - 5x \\ 5x = -2 \pi n- \frac{ 3\pi}{2} \\ x = -\frac{2 \pi n}{5} - \frac{3 \pi}{10}[/latex]