SOLUTION: Solve the equation 2sin^2A = cosA + 2 over the interval [0,2π]

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Question 594162: Solve the equation 2sin^2A = cosA + 2 over the interval [0,2π]
Answer by lwsshak3(11628) About Me  (Show Source):
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Solve the equation 2sin^2A = cosA + 2 over the interval [0,2π]
2sin^2=cos+2
2(1-cos^2)=cos+2
2-2cos^2=cos+2
2cos^2+cos=0
cos(2cos+1)=0
..
cosA=0
A=π/2 and 3π/2
..
2cosA+1=0
cosA=-1/2
A=2π/3 and 5π/3 (in quadrants III and IV where cos<0)