cosx+cos 5x= cos 3x+cos 7x

Решите задачу:

$cosx+cos5x = cos3x + cos7x\\ 2cos\frac{x+5x}{2}cos\frac{x-5x}{2} = 2cos\frac{3x+7x}{2}cos\frac{3x-7x}{2}\\ cos3x*cos2x = cos5x * cos2x\\ cos3x*cos2x - cos5x*cos2x = 0\\ cos2x(cos3x - cos5x) = 0\\ 1)cos2x = 0\ \ \ \ \ \ \ 2)cos3x-cos5x = 0\\ 1)cos2x = 0\\ 2x = \frac{\pi}{2} + \pi*n\\ x = \frac{\pi}{4} + \frac{\pi}{2} *n\\ 2) cos3x-cos5x = 0\\ -2sin\frac{3x-5x}{2}sin\frac{3x+5x}{2}=0\\ 2sinx*sin4x=0\\ sinx=0\ \ \ \ \ \ \ \ \ \ sin4x = 0\\ x = \pi*k \ \ \ \ \ \ \ \ \ x = \frac{\pi}{4}*k$