Sin (4x) + cos (6x) = 0

Sin (4x) + cos (6x) = 0
$sin (4x) + cos (6x) = 0 \\ \\ cos( \frac{ \pi }{2} -4x)+cos(6x)=0 \\ \\2cos( \frac{\frac{ \pi }{2} -4x+6x}{2} })cos( \frac{ \frac{ \pi }{2} -4x-6x}{2} } ) =0 \\ \\ 2cos( \frac{ \pi }{4}+x )cos( \frac{ \pi }{4}-5x )=0 \\ \\ 1) cos( \frac{ \pi }{4}+x )=0 \\ \\ \frac{ \pi }{4} +x = \frac{ \pi }{2} + \pi n \\ \\ x= \frac{ \pi }{2} - \frac{ \pi }{4} + \pi n \\ \\ x = \frac{ \pi }{4} + \pi n$

$2)cos( \frac{ \pi }{4}-5x )=0 \\ \\ cos( 5x - \frac{ \pi }{4} )=0 \\ \\ 5x - \frac{ \pi }{4} = \frac{ \pi }{2} + \pi k \\ \\ 5x = \frac{ \pi }{2} + \frac{ \pi }{4} + \pi k \\ \\ 5x= \frac{3 \pi }{4} + \pi k \\ \\ x = \frac{3 \pi }{20} + \frac{ \pi }{5} k$

Ответ
$x_1 = \frac{ \pi }{4} + \pi n \\ \\ x_2 = \frac{3 \pi }{20} + \frac{ \pi }{5} k$ n,k∈Z