Sqrt2 cos5x+sin3x-sin7x=0

$\sqrt{2} cos5x +sin3x-sin7x = 0$
$sin3x-sin7x = 2sin \frac{3x-7x}{2} cos \frac{3x+7x}{2} = 2sin(-2x)cos5x = -2sin2xcos5$ $x$
$\sqrt{2} cos5x-2sin2xcos5x = 0$
$cos5x( \sqrt{2} - 2sin2x)=0$
$cos5x = 0$
$5x = б \frac{ \pi }{2} +2 \pi n$
$x = б \frac{ \pi }{10} + \frac{2 \pi n}{5}$

$\sqrt{2} - 2sin2x=0$
$- 2sin2x=-\sqrt{2}$
$2sin2x = \sqrt{2}$
$sin2x = \frac{ \sqrt{2} }{2}$
$2x = (-1)^n \frac{ \pi }{4} + \pi n$
$x = (-1)^n \frac{ \pi }{8} + \frac{ \pi n}{2}$

ответ: $x_{1}= б \frac{ \pi }{10} + \frac{2 \pi n}{5}$
$x_{2} = (-1)^n \frac{ \pi }{8} + \frac{ \pi n}{2}$