1. ## Solving an equaion.

Find the zeros of the equation $2cos^2x+cos2x-1 = 0$ between $0$ and $2\pi$.

2. $2\cos^2x+\cos2x-1 = 0$

$2\cos^2x+(2\cos^2x-1)-1 = 0$

$4\cos^2x-2 = 0$

$4\cos^2x = 2$

$\cos^2x = \frac{1}{2}$

$\cos(x) = \frac{1}{\sqrt{2}}$

Should be easy from here, there are 2 solutions

Find the zeros of the equation $2cos^2x+cos2x-1 = 0$ between $0$ and $2\pi$.
$2\cos^2{x} + \cos(2x) - 1 = 0$

note the double angle identity ... $\cos(2x) = 2\cos^2{x} - 1
$

two ways to go from here ...

$(2\cos^2{x} - 1) + \cos(2x) = 0$

$\cos(2x) + \cos(2x) = 0$

$2\cos(2x) = 0$

$2x = \frac{\pi}{2} \, , \, \frac{3\pi}{2} \, , \, \frac{5\pi}{2} \, , \, \frac{7\pi}{2}$

$x = \frac{\pi}{4} \, , \, \frac{3\pi}{4} \, , \, \frac{5\pi}{4} \, , \, \frac{7\pi}{4}$

or ...

$(2\cos^2{x} - 1) + \cos(2x) = 0$

$(2\cos^2{x} - 1) + (2\cos^2{x} - 1) = 0$

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

$\cos^2{x} = \frac{1}{2}$

$\cos{x} = \pm \frac{1}{\sqrt{2}}$

$x = \frac{\pi}{4} \, , \, \frac{3\pi}{4} \, , \, \frac{5\pi}{4} \, , \, \frac{7\pi}{4}$