# How do I solve 2 cos^2 x-sin x+1=0 for 0° <=x<=360°?

Sep 28, 2015

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

#### Explanation:

${\cos}^{2} x = 1 - {\sin}^{2} x$
$2 \left(1 - {\sin}^{2} x\right) - \sin x + 1 = 0$
$2 - 2 {\sin}^{2} x - \sin x + 1 = 0$
$- 2 {\sin}^{2} x - \sin x + 3 = 0$
$2 {\sin}^{2} x + \sin x - 3 = 0$
$2 {\sin}^{2} x - 4 \sin x + 5 \sin x + 2 - 2 - 3 = 0$
$2 {\sin}^{2} x - 4 \sin x + 2 + 5 \sin x - 2 - 3 = 0$
$2 \left({\sin}^{2} x - 2 \sin x + 1\right) + 5 \sin x - 5 = 0$
$2 {\left(\sin x - 1\right)}^{2} + 5 \left(\sin x - 1\right) = 0$
$\left(\sin x - 1\right) \left(2 \left(\sin x - 1\right) + 5\right) = 0$
$\left(\sin x - 1\right) \left(2 \sin x - 2 + 5\right) = 0$
$\left(\sin x - 1\right) \left(2 \sin x + 3\right) = 0$
$\sin x - 1 = 0 \vee 2 \sin x + 3 = 0$
$\sin x = 1 \vee \sin x = - \frac{3}{2}$

$x = \frac{\pi}{2}$ is a solution of the first equation.
$- 1 \le \sin x \le 1$ so the second equation has no solution.

Finally, $x = \frac{\pi}{2}$ is a solution.