Question

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

Answer 1

just want to know how to start

Answer 2

hint:

sin^2 = 1 - cos^2

Answer 3

i still dont get it can you write the first part for me

Answer 4

2sin^2 x -cos x =1

2(1 - cos^2x) -cos x =1

2 - 2cos^2x -cos x =1

-2cos^2x - cosx + 2 = 1

-2cos^2x - cosx + 2 - 1 = 0

-2cos^2x - cosx + 1 = 0

-2z^2 - z + 1 = 0

now solve for z

Answer 5

thnks

Answer 6

you're welcome