Need help with trigonometric equations please.

**derek1008** · November 11th 2012, 08:18 AM

How would I solve the trig equation 5 cos(2x)=2 ?????????????

**richard1234** · November 11th 2012, 09:11 AM

$\cos 2x = \frac{2}{5}$

What angles within the unit circle have a cosine of 2/5?

**skeeter** · November 11th 2012, 09:17 AM

assuming solutions in the interval $0 < x < 2\pi$ ...

$0 < x < 2\pi \implies 0 < 2x < 4\pi$

let $t = 2x$ , so $0 < t < 4\pi$

$5 \cos{t} = 2$

$\cos{t} = \frac{2}{5}$

since $\cos{t} > 0$ , $t$ is an angle in quads I or IV

quad I solutions for $t$ , (two) ...

$t = \arccos\left(\frac{2}{5}\right)$ and $t = 2\pi + \arccos\left(\frac{2}{5}\right)$

so $x = \frac{1}{2}\arccos\left(\frac{2}{5}\right)$ and $x = \pi + \frac{1}{2}\arccos\left(\frac{2}{5}\right)$

quad IV solutions (two) ...

$t = 2\pi - \arccos\left(\frac{2}{5}\right)$ and $t = 4\pi - \arccos\left(\frac{2}{5}\right)$

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

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Need help with trigonometric equations please.

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