Question

Find all solutions of the equation 2sin^2 x -cos x =1 in the interval [0,2pi).

Answer 1

2sin^2x-cosx=1
or 2(1-cos^2x)-cos x=1
2cos^2x+cosx-1=0
(cosx+1)(2cosx-1)=0
so cosx=-1 and cosx=1/2

Answer 2

thnk u

Answer 3

so the second line what happened to the 2 when distributed isnt it 2-2cos^2x-cos x =1

Answer 4

Its a quadratic in equation in cos(x)