22cos^2x+8sinx*cosx=7

Решите задачу:

$22\cos^2x+8\sin x\cos x=7\\22\cos^2x+8\sin x\cos x-7\cos^2x-7\sin^2x=0\\15\cos^2x-15\sin^2x+8\sin^2x+8\sin x\cos x=0\\15(\cos x-\sin x)(\cos x+\sin x)+8\sin x(\sin x+\cos x)=0\\(\cos x+\sin x)(15\cos x-15\sin x+8\sin x)=0\\(\cos x+\sin x)(15\cos x-7\sin x)=0\\\begin{cases}\cos x+\sin x=0\\15\cos x-7\sin x=0\end{cases}\Rightarrow\begin{cases}tgx=-1\\tgx=\frac{15}7\end{cases}\Rightarrow\begin{cases}x=\frac{3\pi}4+\pi k\\x=arctg\left(\frac{15}7\right)+\pi k\end{cases}$