Допишите до формулы пожалуйста или решите sinx+cosx=√2sin7x

$\sin(x) + \cos(x) = \\ = \sqrt{2} ( \frac{ \sqrt{2} }{2} \sin(x) + \frac{ \sqrt{2} }{2} \cos(x) ) = \\ = \sqrt{2} ( \cos( \frac{\pi}{4} ) \sin(x) + \sin( \frac{\pi}{4} ) \cos(x) ) = \\ = \sqrt{2} \sin(x + \frac{\pi}{4} ) \\$
поэтому наше уравнение равносильно следующему:
$\sqrt{2} \sin(x + \frac{\pi}{4} ) = \sqrt{2} \sin(7x)$
$\sqrt{2} ( \sin(x + \frac{\pi}{4} ) - \sin(7x) ) = 0 \\ ( \sin(x + \frac{\pi}{4} ) - \sin(7x) ) = 0 \\ 2 \sin( \frac{x + \frac{\pi}{4} - 7x}{2} ) \cos( \frac{x + \frac{\pi}{4} + 7x}{2}) = 0 \\ \sin( \frac{\pi}{8} - 3x ) \cos( \frac{\pi}{8} + 4x) = 0 \\ - \sin(3x - \frac{\pi}{8} ) \cos(4x + \frac{\pi}{8} ) = 0 \\ \\ \sin(3x - \frac{\pi}{8} ) = 0 \\ 3x - \frac{\pi}{8} = \pi k \\ x = \frac{\pi}{24} + \frac{\pi k}{3} (k∈Z)\\ \\ \cos(4x + \frac{\pi}{8} ) = 0 \\ 4x + \frac{\pi}{8} = \frac{\pi}{2} + \pi \: n \\ 4x = \frac{3\pi}{8} + \pi n \\ x = \frac{3\pi}{32} + \frac{\pi n}{4} (n∈Z)$
Ответ

$x = \frac{\pi}{24} + \frac{\pi k}{3} (k∈Z) \\ \\ x = \frac{3\pi}{32} + \frac{\pi n}{4} (n∈Z)$