Решите уравнение cosxsin9x=cos3xsin7x

$sin \alpha cos \beta = \frac{sin( \alpha + \beta)+sin( \alpha - \beta ) }{2}$

$\frac{1}{2}( sin(9x+x)+sin(9x-x))= \frac{1}{2} (sin(3x+7x)+sin(7x-3x)), \\ sin(9x+x)+sin(9x-x)=sin(3x+7x)+sin(7x-3x), \\ sin10x +sin8x=sin10x+sin4x, \\ sin8x=sin4x \\ 2sin4xcos4x=sin4x \\ sin4x=0, 2cos4x=1 \\ 4x= \pi n, \\ x= \frac{ \pi }{4}n \\ cos4x= \frac{1}{2} \\ 4x=+- \frac{ \pi }{3}+2 \pi k \\ x=+-\frac{\pi}{12}+\frac{\pi k}{2}$
Ответ:
х∈{πn/4;-π/12+πn/2; π/12+πn/2}, n∈Z