# How do you solve 4cos^2x-1=0?

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Steve M Share
Oct 21, 2016

$x = \pm \frac{4 \pi}{3} , \pm \frac{2 \pi}{3} , \pm \frac{5 \pi}{3} , \pm \frac{\pi}{3}$

#### Explanation:

This is a trivial quadratic equation in $\cos x$

$4 {\cos}^{2} x - 1 = 0$
$\therefore 4 {\cos}^{2} x = 1$
$\therefore {\cos}^{2} x = \frac{1}{4}$
$\therefore \cos x = \pm \frac{1}{2}$

In order to solve this you need to be familiar with the graphs of the trig functions along with common values:

The range of soutions is not specified in the question, so lets assume we want a solutio in radians, and $- 2 \pi \le x \le 2 \pi$

Consider, $\cos = \frac{1}{2} \implies x = - 2 \pi + \frac{\pi}{3} , - \frac{\pi}{3} , \frac{\pi}{3} , 2 \pi - \frac{\pi}{3}$
$\therefore x = \frac{- 5 \pi}{3} , \frac{- \pi}{3} , \frac{\pi}{3} , \frac{5 \pi}{3}$

Consider, $\cos = - \frac{1}{2} \implies x = - 2 \pi + \frac{2 \pi}{3} , \frac{- 2 \pi}{3} , \frac{2 \pi}{3} , 2 \pi - \frac{2 \pi}{3}$
$\therefore x = \frac{- 4 \pi}{3} , \frac{- 2 \pi}{3} , \frac{2 \pi}{3} , \frac{4 \pi}{3}$

So the solutions are:
$\therefore x = \pm \frac{4 \pi}{3} , \pm \frac{2 \pi}{3} , \pm \frac{5 \pi}{3} , \pm \frac{\pi}{3}$

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