Sin^2x - 3sinx cosx = 1 Срочно!!!!

Решите задачу:

$\sin^2x-3\sin x\cos x=1\\ \\ \sin^2x-3\sin x\cos x=\sin^2x+\cos^2x\\ \\ \cos^2x+3\sin x\cos x=0\\ \\ \cos x(\cos x+3\sin x)=0\\ \\ \left[\begin{array}{ccc}\cos x=0\\ tgx=-1/3\end{array}\right~~~\Rightarrow~~~~ \left[\begin{array}{ccc}x= \frac{\pi}{2}+ \pi n,n \in \mathbb{Z}\\ \\ x=-arctg1/3+ \pi n,n \in \mathbb{Z} \end{array}\right$