# How do you solve 3cos^2 x - 5cos x -1= 3 sin^2 x?

Sep 1, 2015

$x = 2 n \pi \pm 2 \frac{\pi}{3}$

The description looks long, but I've written every single step so that anyone can understand!

#### Explanation:

3cos^2x−5cosx−1=3sin^2x

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

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

$6 {\cos}^{2} x - 5 \cos x - 4 = 0$

from here its just a quadratic equation. you divide it to its factors.

$\left(3 \cos x - 4\right) \left(2 \cos x + 1\right) = 0$

$3 \cos x - 4 = 0$ or $2 \cos x + 1 = 0$
from here u get two possibilities

$\cos x = \frac{4}{3}$ or $\cos x = - \frac{1}{2}$
the first answer cannot be true because the cosines should be between -1 and +1
if u remember the cosine curve u should be able to understand what im talking about.

so we are left with the latter answer

$\cos x = - \frac{\pi}{3}$
$\cos x = \left(\pi - \frac{\pi}{3}\right)$ you dont really have to show this step.. u can do it in your mind ;)
$\cos x = 2 \frac{\pi}{3}$

according to the trig equations $\theta = 2 n \pi \pm \theta$
just replace the correct places of this equations with our answer
$x = 2 n \pi \pm 2 \frac{\pi}{3}$