Question 553772
Solve ( 2sin^2x - cos^2x - 2 = 0 ) for ( 0° &#8804; &#952; < 360° )
**
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