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

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

2 Answers
Feb 15, 2018

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)=1/2and 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.

Feb 16, 2018

(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