# 2cos^2x-sinx=-1?

May 17, 2018

pi/6; (5pi)/6; (3pi)/2

#### Explanation:

Solve the equation:
$2 {\cos}^{2} x - \sin x + 1 = 0$
Replace in the equation ${\cos}^{2} x$ by $\left(1 - {\sin}^{2} x\right)$ -->
$2 - 2 {\sin}^{2} x - \sin x - 1 = 0$
Solve this quadratic equation for sin x -->
$- 2 {\sin}^{2} x - \sin x + 1 = 0$
Since a - b + c = 0, use shortcut. The 2 real roots are:
sin x = -1 and $\sin x = - \frac{c}{a} = \frac{1}{2}$
a. sin x = - 1
Unit circle gives --> $x = \frac{3 \pi}{2} + 2 k \pi$
b. $\sin x = \frac{1}{2}$
Trig table and unit circle give 2 solutions for x -->
$x = \frac{\pi}{6}$, and $x = \frac{5 \pi}{6}$