Question 1060046
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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
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