Question

sin x^2 +sin x - 2=0​

Answer 1

$Sin^2x + Sinx -2 = 0$

$Sin^2x - Sinx + 2Sinx - 2 = 0$

$Sinx(Sinx-1)+2(Sinx-1) = 0$

$(Sinx+2)(Sinx-1) = 0$

$Sinx = -2$   OR   $Sinx = 1$

Range of $Sinx$ = $[-1,1]$

Hence $Sinx = -2$ is rejected

$Sinx = 1$

$Sinx = Sin\frac{\pi }{2}$

$x = \frac{\pi }{2}$

HOPE IT HELPS !!