SOLUTION: Find all solutions of the following equation: 2 cos^2(x) - 3 cos(x) = 2 sin(x)cos(x) - 3 sin(x) I am absolutely clueless as to how to solve this. Please help!!

Click here to see ALL problems on Trigonometry-basics

Question 1060046: Find all solutions of the following equation:
2 cos^2(x) - 3 cos(x) = 2 sin(x)cos(x) - 3 sin(x)
I am absolutely clueless as to how to solve this. Please help!!
Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website!

First get everything to one side

2 cos^2(x) - 3 cos(x) = 2 sin(x)cos(x) - 3 sin(x)
2 cos^2(x) - 3 cos(x) - 2 sin(x)cos(x) + 3 sin(x) = 0

Now separate the terms into two groups and factor by grouping

2 cos^2(x) - 3 cos(x) - 2 sin(x)cos(x) + 3 sin(x) = 0
(2 cos^2(x) - 3 cos(x)) + (-2 sin(x)cos(x) + 3 sin(x)) = 0
cos(x)(2 cos(x) - 3) + (-2 sin(x)cos(x) + 3 sin(x)) = 0
cos(x)(2 cos(x) - 3) -sin(x)(2 cos(x) - 3) = 0
(cos(x) - sin(x))(2 cos(x) - 3) = 0

After everything is factored, you can set each factor equal to 0 and solve for x

Set the first factor equal to 0
cos(x) - sin(x) = 0
cos(x) = sin(x)
Then use the unit circle to determine that the solutions occur at x = pi/4 and x = 5pi/4

Set the other factor equal to 0 and solve for x
2 cos(x) - 3 = 0
2 cos(x) = 3
cos(x) = 3/2
cos(x) = 1.5 ... no solutions
Recall that the range of cosine has 1 as the largest value. So 1.5 is not possible.

The only two solutions in the interval [0,2pi) are
x = pi/4
x = 5pi/4

Tack on "+2pi*n" to handle the rest of the solutions

So the entire solution set is
x = pi/4 + 2pi*n
x = 5pi/4 + 2pi*n
where n is any integer