Question 448352
can you please help me solve this trigonometric equation: 
2sin^2x-cosx-1=0
..
for 0<x<2&#960;
2sin^2x-cosx-1=0
2(1-cos^2x)-cosx-1=0
2-2cos^2x-cosx-1=0
-2cos^2x-cosx+1=0
2cos^2x+cosx-1=0
(2cosx-1)(cosx+1)=0
2cosx-1=0
cosx=1/2
x=&#960;/3,5&#960;/3
..
cosx+1=0
cosx=-1
x=&#960;
..
ans:
x=&#960;, &#960;/3, 5&#960;/3