Необходимо решить уравнение: (1/tg^2 x) - (1/sin x) - 1 = 0

Решите задачу:

$\frac1{tg^2x} -\frac1{\sin x} - 1 = 0\\\frac1{\frac{\sin^2x}{\cos^2x}}-\frac1{\sin x}-1=0\\\frac{\cos^2x}{\sin^2x}-\frac1{\sin x}-1=0\\\frac{1-\sin^2x}{\sin^2x}-\frac1{\sin x}-1=0\\\frac1{\sin^2x}-1-\frac1{\sin x}-1=0\\\frac1{\sin^2x}-\frac1{\sin x}-2=0\\\sin x\neq0\Rightarrow x\neq\pi n,\;n\in\mathbb{Z}\\\frac1{\sin x}=t,\;\frac1{\sin^2x}=t^2,\;t\neq0\\t^2-t-2=0\\D=1+4\cdot2=9\\t_1=-1,\;t_2=2\\\frac1{\sin x}=-1\Rightarrow\sin x=-1\Rightarrow x=\pi+2\pi n,\;n\in\mathbb{Z}$
$\frac1{\sin x}=2\Rightarrow\sin x=\frac12\Rightarrow x=\frac\pi6+2\pi n,\;n\in\mathbb{Z}$