# How do you find a numerical value of one trigonometric function of x given cos^2x+2sinx-2=0?

Nov 14, 2016

$\therefore x = \frac{\pi}{2} + 2 \pi n$

#### Explanation:

${\cos}^{2} x + 2 \sin x - 2 = 0$

$\left(1 - {\sin}^{2} x\right) + 2 \sin x - 2 = 0$

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

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

Now factor.

$\left(\sin x - 1\right) \left(\sin x - 1\right) = 0$

${\left(\sin x - 1\right)}^{2} = 0$

$\sin x - 1 = 0$

$\sin x = 1$

$x = {\sin}^{-} 1 1$

$\therefore x = \frac{\pi}{2} + 2 \pi n$