Question

How to solve this equation `1 + cos theta = 2 sin^2 theta` over the domain `0 <= theta <= 2pi` ( Solve for `theta` )?

Answer 1

Solution: `theta= pi/3 , theta= pi , theta= (5pi)/3`

Explanation:

`1+cos theta = 2 sin^2 theta or 1+cos theta = 2(1- cos^2 theta) or 2 cos^2 theta +cos theta -1= 0 or 2 cos^2 theta +2cos theta -cos theta -1= 0 or 2cos theta(cos theta+1) -1(cos theta+1) =0 or (2cos theta-1)(cos theta+1) =0 :.` Either `2cos theta - 1 =0 or cos theta +1 =0`

Case 1) `2cos theta - 1 =0 or 2cos theta =1 or cos theta = 1/2 ; cos (pi/3) = 1/2 :. theta = pi/3 ; cos (2pi-pi/3) =1/2 :. theta = (5pi)/3`

Case 2) `cos theta +1 =0 or cos theta = -1 ; cos (pi) = -1 :. theta = pi`

Solution: `theta= pi/3 , theta= pi , theta= (5pi)/3` [Ans]