Решите пожалуйста, 2sin²x-3sin x+1=0

Сделаем замену
$\sin(x) = t \\ 2 {t}^{2} - 3t + 1 = 0 \\ d = 9 - 8 = 1 \\ x1 =( 3 + 1) \div4 = 1 \\ x2 = (3 - 1) \div 4 = 0.5$
$\sin(x) = 1 \\ x = \frac{\pi}{2} + 2\pi k$
$\sin(x) = \frac{1}{2} \\ x = \frac{\pi}{3} + 2\pi k \\ and \\ x = \frac{2\pi}{3} + 2\pi k$
Ответ.

$x = \frac{\pi}{2} + 2\pi k$
$x = \frac{\pi}{3} + 2\pi k$
$x = \frac{2\pi}{3} + 2\pi k$