Question 252486
Here is the original problem:
(i) 2sin^2x+3cosx=3
We have an identity for sin^2(x). 
Sin^2(x) + Cos^2(x) = 1.
Solving for Sin^2(x), we get
sin^2(x) = 1 - cos^2(x)
Now, by substitution into (i), we get
2(1 - cos^2(x)) + 3cos(x) - 3 = 0
2cos^2(x) - 3cos(x) + 1 = 0.
Factoring, we get
(2cos(x) - 1)(cos(x) - 1) = 0.
Solve each parenthesis for x.
(2cos(x) - 1) = 0
cos(x) = 1/2. 
x = 60 degrees or 300 degrees ; radians = pi/3, 5pi/3
(cos(x) - 1) = 0
cos(x) = 1
x = 0 degrees ; 0 radians.
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So, your answers are: 0, 60 degrees, 300 degrees.