Уравнение тригонометричесеое........... хелп:) 16 задание!!!!

$2sinx - cos^{2} x -2 = 0$

$2sinx - (1- sin^{2}x ) -2 = 0$

$2 sinx -1 + sin^{2} x -2 = 0$

$sin^{2} x + 2sinx - 3 = 0$

sinx = t

$t^{2} + 2t - 3 = 0$

0" alt="D = b^{2} - 4ac = 4 +4*3 = 4 + 12 = 16 = 4^{2} >0" align="absmiddle" class="latex-formula">

$x1 = \frac{-2+4}{2} = \frac{2}{2} = 1$
$x2 = \frac{-2-4}{2} = \frac{-6}{2} = -3$ (корень не подходит)

sinx= 1
$x = \frac{ \pi }{2} + 2 \pi k$