Question 731700
cos x + 1 = &#8730;3 sin x where 0 &#8804; x < 2&#960;
cos x + 1 = &#8730;3 sin x
square both sides
cos^2x+2cosx+1=3sin^2x
cos^2x+2cosx+1=3(1-cos^2x)
cos^2x+2cosx+1=3-3cos^2x)
4cos^2x+2cosx-2=0
(4cosx-2)(cosx+1)=0
..
4cosx-2=0
cosx=1/2
x=&#960;/3, 5&#960;/3 (In quadrants I and IV where cos>0)
or 
cosx+1=0
cosx=-1
x=&#960;