Question 610983
Solve algebraically for all values of theta in the interval 0 degrees is < or equal to theta which is < 360 degrees. Express your answers to the nearest degree. 
2sin^2 theta-4sin theta = cos ^2 theta-2
**
2sin^2x-4sinx=cos^2x-2
2sin^2x-4sinx=1-sin^2x-2
3sin^2x-4sinx+1=0
(3sinx-1)(sinx-1)=0
..
3sinx-1=0
sinx=1/3
x&#8776;19º
or
sinx-1=0
sinx=1
x=90º
ans:
Theta&#8776;19º and =90º