# How do I solve 1=2cos(3X-5pi) for (0,2pi)?

## $1 = 2 \cos \left(3 x - 5 \pi\right)$ for $\left(0 , 2 \pi\right]$

Feb 15, 2018

The values of x being $\frac{16 \pi}{9} \mathmr{and} \frac{20 \pi}{9}$ form the valid solutions.

#### Explanation:

Given:
$1 = 2 \cos \left(3 x - 5 \pi\right)$
for $\left(0 , 2 \pi\right)$
Interchanging lhs and rhs
$2 \cos \left(3 x - 5 \pi\right) = 1$

Dividing by 2
$\cos \left(3 x - 5 \pi\right) = \frac{1}{2}$
We know that $\cos \left(\frac{\pi}{6}\right) = \frac{1}{2}$ is the fundamental solution
Comparing
$\cos \left(3 x - 5 \pi\right) = \frac{1}{2}$and $\cos \left(\frac{\pi}{3}\right) = \frac{1}{2}$
we have, $3 x - 5 \pi = \frac{\pi}{3} , 2 \pi - \frac{\pi}{3}$ for the range $\left(0 , 2 \pi\right)$

$3 x - 5 \pi = \frac{\pi}{3}$
Adding $5 \pi$
$3 x = 5 \pi + \frac{\pi}{3} = \frac{31 \pi}{6}$
Dividing by 3
$x = \left(\frac{1}{3}\right) \frac{16 \pi}{3} = \frac{16 \pi}{9}$

$3 x - 5 \pi = 2 \pi - \frac{\pi}{3} = \frac{5 \pi}{3}$
Adding $5 \pi$
$3 x = 5 \pi + \frac{5 \pi}{3} = \frac{20 \pi}{3}$
Dividing by 3
$x = \frac{20 \pi}{9}$

The values of x being $\frac{16 \pi}{9} \mathmr{and} \frac{20 \pi}{9}$ form the valid solutions.

Feb 16, 2018

(2pi)/9; (4pi)/9

#### Explanation:

$2 \cos \left(3 x - 5 \pi\right) = 2 \cos \left(3 x - \pi\right) = 1$
$\cos \left(3 x - \pi\right) = \frac{1}{2}$
Trig table and unit circle give 2 solutions:
$3 x - \pi = \pm \frac{\pi}{3}$
a. $3 x - \pi = \frac{\pi}{3}$ --> $3 x = \pi + \frac{\pi}{3} = \frac{4 \pi}{3}$
$x = \frac{4 \pi}{9}$
b. $3 x - \pi = - \frac{\pi}{3}$ --> $3 x = \pi - \frac{\pi}{3} = \frac{2 \pi}{3}$
$x = \frac{2 \pi}{9}$