# Need help with trigonometric equations please.

• Nov 11th 2012, 08:18 AM
derek1008
Need help with trigonometric equations please.
How would I solve the trig equation 5 cos(2x)=2 ?????????????(Headbang)
• Nov 11th 2012, 09:11 AM
richard1234
Re: Need help with trigonometric equations please.
$\displaystyle \cos 2x = \frac{2}{5}$

What angles within the unit circle have a cosine of 2/5?
• Nov 11th 2012, 09:17 AM
skeeter
Re: Need help with trigonometric equations please.
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 $\displaystyle x = \pi + \frac{1}{2}\arccos\left(\frac{2}{5}\right)$

$\displaystyle t = 2\pi - \arccos\left(\frac{2}{5}\right)$ 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 $\displaystyle x = 2\pi - \frac{1}{2}\arccos\left(\frac{2}{5}\right)$