Question

how can i find all solutions in the interval [0, 2pi) from 2sin^2(x)-cos(x)-1=0

Answer 1

there is a typo... its actually, 2sin^2(x)-cos(x)=1

Answer 2

sub.s in 1-cos^2(x) for sin^2(x)
you'll have a quadratic in cos(x)

Answer 3

The typo and your corrected equation are the same, subtract 1 from both sides from the corrected equation and you'd get the typo.