Решите уравнение 5/sin^2 X

Решите задачу:

$\frac{5}{ \sin {}^{2} (x) } - \frac{3}{ \sin(x) } - 2 = 0 \\ \frac{5 - 3 \sin(x) - 2 \sin {}^{2} (x) }{ \sin(x) } = 0 \\ \sin(x) ≠0 \\ x≠\pi \: k \: \\ - 2 \sin {}^{2} (x) - 3 \sin(x) + 5 = 0/ \times ( - 1) \\ 2 \sin {}^{2} (x) + 3 \sin(x) - 5 = 0 \\ \sin(x) = t \: \: \: - 1 \leqslant t \leqslant 1 \\ 2 {t}^{2} + 3t - 5 = 0 \\ x = \frac{ -3 ± \sqrt{9 + 4 \times 5 \times 2} }{4} = \frac{ - 3±7}{4} = - \frac{5}{2} ;1 \\ \sin(x) = 1 \\ x = \frac{\pi}{2} + 2\pi \: k \:; k\in\mathbb Z \\$
$\pi \leqslant \frac{\pi}{2} + 2\pi \: k \leqslant \frac{5\pi}{2} \\ \frac{\pi}{2} \leqslant 2\pi \: k \leqslant 2\pi \\ \frac{1}{4} \leqslant k \leqslant 1 \\ k = 1 \\ \frac{\pi}{2} + 2\pi = \frac{5\pi}{2}$