Question 926043
Solve the following, where 0 < (theta) < 2(pi) (the signs are both smaller than or equal to)
2Sin^2(theta) = 2 + 3Cos(theta)
***
2sin^2(x)=2+3cosx
2(1-cos^2(x)=2+3cosx
2-2cos^2(x)=2+3cosx
2cos^2(x)+3cosx=0
cosx(2cosx+3)=0
cosx=0
x=&#960;/2, 3&#960;/2
..
2cosx+3=0
cosx=-3/2
no solution:(-1 &#8804; cosx &#8804; 1)