SOLUTION: Find all solution of the equation in the interval [0,2pi) 4cos(x) = ((sin^2)(x)) + 4

Question 980947: Find all solution of the equation in the interval [0,2pi)
4cos(x) = ((sin^2)(x)) + 4
Found 2 solutions by Alan3354, Boreal:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find all solution of the equation in the interval [0,2pi)
4cos(x) = sin^2(x) + 4
4cos(x) = 1 - cos^2(x) + 4
cos^2 + 4cos - 5 = 0
(cos + 5)*(cos - 1) = 0
cos = -5 (Ignore)
=================
cos = 1
x = 0
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
4cos(x) = ((sin^2)(x)) + 4
4cos x =(1-cos^2 (x))+4
cos^2x+4 cos x-5=0
(cos (x) +5)(cos (x)-1)=0
cos x=1
x=0 on the interval [0,2pi)
at 0 degrees, cos x=1
4=0+4

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