Find all solutions for 2/3 sin^2 x - cos x = 1 on the interval [0,2pi)

Answer 1

2/3sin^2x-cosx=1 can be re-written as

2/3(1-cos^2x)-cosx=1.

Multiplying each term by 3 gives 2(1-cos^2x)-3cosx=3.

This gives 2-2cos^2x-3cosx=3, or -2cos^2x-3cosx=1.

Multiplying each term by -1 gives 2cos^2x+3cosx=-1.

Rearranging to make the right side zero gives 2cos^2x+3cosx+1=0

This factors into (2cosx+1)(cosx+1)=0.

Setting each factor=0 gives 2cosx=-1,cosx=-1/2, and x=2pi/3 and 4pi/3 and then cosx=-1, x=pi.

So, the solutions on the interval [0,2pi) are 2pi/3, pi and 4pi/3.