SOLUTION: Solve the following, where 0 < (theta) < 2(pi) (the signs are both smaller than or equal to) 2Sin^2(theta) = 2 + 3Cos(theta)

Question 926043: Solve the following, where 0 < (theta) < 2(pi) (the signs are both smaller than or equal to)
2Sin^2(theta) = 2 + 3Cos(theta)

Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Solve the following, where 0 < (theta) < 2(pi) (the signs are both smaller than or equal to)
2Sin^2(theta) = 2 + 3Cos(theta)
***
2sin^2(x)=2+3cosx
2(1-cos^2(x)=2+3cosx
2-2cos^2(x)=2+3cosx
2cos^2(x)+3cosx=0
cosx(2cosx+3)=0
cosx=0
x=π/2, 3π/2
..
2cosx+3=0
cosx=-3/2
no solution:(-1 ≤ cosx ≤ 1)
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