Решите, даю 22 бала cosx+7sinx-5=0

Решите задачу:

$cosx+7sinx-5=0\\ Acosx+Bcosx=Csin(x+t);\ C=\sqrt{A^2+B^2}\\ C=\sqrt{1^2+7^2}=\sqrt{50}=5\sqrt2\\ \frac{1}{5\sqrt2}cosx+\frac{7}{5\sqrt2}sinx-\frac{5}{5\sqrt2}=0\\ \frac{\sqrt2}{10}cosx+\frac{7\sqrt2}{10}sinx=\frac{\sqrt2}{2}\\ t=arcsin\frac{\sqrt2}{10}\\ sint*cosx+cost*sinx=\frac{\sqrt2}{2}\\ sin(x+t)=\frac{\sqrt2}{2}\\ |x+t=\frac{\pi}{4}+2\pi k\\ |x+t=\frac{3\pi}{4}+2\pi k\\ \\ |x=\frac{\pi}{4}-arcsin\frac{\sqrt2}{10}+2\pi k\\ |x=\frac{3\pi}{4}-arcsin\frac{\sqrt2}{10}+2\pi k$