This is what the problem says:
solve the equation for solutions in the interval [0,2pi]
showing all steps would be appreciated.
2sin²(x) - cos(x) - 1 = 0
I solved it this way:
Since $\displaystyle \sin^2(x)+\cos^2(x)=1$, then $\displaystyle \sin^2(x)=1-\cos^2(x)$
substitute in your equation and you get
$\displaystyle 2(1-\cos^2(x))-\cos(x)-1=0$
using distributive and reordering terms you get
$\displaystyle -2 \cos^2(x)-\cos(x)+1=0$
Now you have to make a change of variables, so let $\displaystyle u=\cos(x)$, then $\displaystyle x=\cos^{-1}(u)$ (you will need this later)
and substituting in this last equation you get
$\displaystyle -2u^2-u+1=0$
which has roots $\displaystyle u_1=-1$ and $\displaystyle u_2=1/2$
Then $\displaystyle x_1=\cos^{-1}(-1)=\pi$ and $\displaystyle x_2=\cos^{-1}(1/2)=\pi/3$ or $\displaystyle 5\pi/3\$
Those are the 3 solutions