Question

Sqrt2 cos5x+sin3x-sin7x=0

Answer 1

$sqrt{2} cos5x +sin3x-sin7x = 0$
$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: $x_{1}= б frac{ pi }{10} + frac{2 pi n}{5}$
$x_{2} = (-1)^n frac{ pi }{8} + frac{ pi n}{2}$