Question

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

`1=2cos(3x-5pi)` for `(0,2pi]`

Answer 1

The values of x being `(16pi)/9 and (20pi)/9` form the valid solutions.

Explanation:

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

Dividing by 2
`cos(3x-5pi)=1/2`
We know that `cos(pi/6)=1/2` is the fundamental solution
Comparing
`cos(3x-5pi)=1/2`and `cos(pi/3)=1/2`
we have, `3x-5pi=pi/3,2pi-pi/3` for the range `(0,2pi)`

`3x-5pi=pi/3`
Adding `5pi`
`3x=5pi+pi/3=(31pi)/6`
Dividing by 3
`x=(1/3)(16pi)/3=(16pi)/9`

`3x-5pi=2pi-pi/3=(5pi)/3`
Adding `5pi`
`3x=5pi+(5pi)/3=(20pi)/3`
Dividing by 3
`x=(20pi)/9`

The values of x being `(16pi)/9 and (20pi)/9` form the valid solutions.

Answer 2

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

Explanation:

`2cos (3x - 5pi) = 2cos (3x - pi) = 1`
`cos (3x - pi) = 1/2`
Trig table and unit circle give 2 solutions:
`3x - pi = +- pi/3`
a. `3x - pi = pi/3` --> `3x = pi + pi/3 = (4pi)/3`
`x = (4pi)/9`
b. `3x - pi = - pi/3` --> `3x = pi - pi/3 = (2pi)/3`
`x = (2pi)/9`