Question

Solve the trigonometric equation 3-3cosx=2 sin^2 x

Answer 1

= 3 - 3cos x - 2(1- cos^2 x)
= 2cos^2 x - 3cos x + 1
=(cos x -1)(cosx - 1/2)
   now cos x = 1    or cos x= 1/2 
   draw a graph we get = + ve values
  hence .. 0 ,180,360 and 60,240, degrees in range x<2pi