SOLUTION: Solve ( 2sin^2x - cos^2x - 2 = 0 ) for ( 0° &#8804; &#952; < 360° ).

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Question 553772: Solve ( 2sin^2x - cos^2x - 2 = 0 ) for ( 0° ≤ θ < 360° ).
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve ( 2sin^2x - cos^2x - 2 = 0 ) for ( 0° ≤ θ < 360° )
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2sin^2x-cos^2x-2=0
2sin^2x-(1-sin^2x)-2=0
2sin^2x-1+sin^2x-2=0
3sin^2x-3=0
3sin^2x=3
sin^2x=1
sinx=±1
x=90º and 270º