# How do you find all solutions of 2cos^2x-sinx-1=0?

Feb 9, 2015

$2 {\cos}^{2} x - \sin x - 1 = 0$ for
$x \in \left\{\frac{3 \pi}{2} + 2 n \pi , \frac{\pi}{6} + 2 n \pi , \frac{5 \pi}{6} + 2 n \pi\right\}$ where $n \in \mathbb{Z}$

Solve : $2 {\cos}^{2} x - \sin x - 1 = 0$ (1)

First, replace ${\cos}^{2} x$ by $\left(1 - {\sin}^{2} x\right)$

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

Call $\sin x = t$, we have:
$- 2 {t}^{2} - t + 1 = 0$.
This is a quadratic equation of the form $a {t}^{2} + b t + c = 0$ that can be solved by shortcut:
$t = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
or factoring to $- \left(2 t - 1\right) \left(t + 1\right) = 0$

One real root is ${t}_{1} = - 1$ and the other is ${t}_{2} = \frac{1}{2}$.

Next solve the 2 basic trig functions:
${t}_{1} = \sin {x}_{1} = - 1$
$\rightarrow$ ${x}_{1} = \frac{\pi}{2} + 2 n \pi$ (for $n \in \mathbb{Z}$)
and
${t}_{2} = \sin {x}_{2} = \frac{1}{2}$
$\rightarrow$ ${x}_{2} = \frac{\pi}{6} + 2 n \pi$
or
$\rightarrow$ ${x}_{2} = \frac{5 \pi}{6} + 2 n \pi$

Check with equation (1):
cos (3pi/2) = 0; sin (3pi/2) = -1
$x = 3 \frac{\pi}{2} \rightarrow 0 + 1 - 1 = 0$ (correct)
cos (pi/6) = (sqrt 3)/2 rarr 2*cos^ 2(pi/6) = 3/2; sin (pi/6) = 1/2.
$x = \frac{\pi}{6} \rightarrow \frac{3}{2} - \frac{1}{2} - 1 = 0$ (correct)