How would I solve the trig equation 5 cos(2x)=2 ?????????????
assuming solutions in the interval $\displaystyle 0 < x < 2\pi$ ...
$\displaystyle 0 < x < 2\pi \implies 0 < 2x < 4\pi$
let $\displaystyle t = 2x$ , so $\displaystyle 0 < t < 4\pi$
$\displaystyle 5 \cos{t} = 2$
$\displaystyle \cos{t} = \frac{2}{5}$
since $\displaystyle \cos{t} > 0$ , $\displaystyle t$ is an angle in quads I or IV
quad I solutions for $\displaystyle t$ , (two) ...
$\displaystyle t = \arccos\left(\frac{2}{5}\right)$ and $\displaystyle t = 2\pi + \arccos\left(\frac{2}{5}\right)$
so $\displaystyle x = \frac{1}{2}\arccos\left(\frac{2}{5}\right)$ and $\displaystyle x = \pi + \frac{1}{2}\arccos\left(\frac{2}{5}\right)$
quad IV solutions (two) ...
$\displaystyle t = 2\pi - \arccos\left(\frac{2}{5}\right)$ and $\displaystyle t = 4\pi - \arccos\left(\frac{2}{5}\right)$
so $\displaystyle x = \pi - \frac{1}{2}\arccos\left(\frac{2}{5}\right)$ and $\displaystyle x = 2\pi - \frac{1}{2}\arccos\left(\frac{2}{5}\right)$