# Math Help - trig help

1. ## trig help

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

2. I solved it this way:
Since $\sin^2(x)+\cos^2(x)=1$, then $\sin^2(x)=1-\cos^2(x)$

substitute in your equation and you get
$2(1-\cos^2(x))-\cos(x)-1=0$

using distributive and reordering terms you get
$-2 \cos^2(x)-\cos(x)+1=0$

Now you have to make a change of variables, so let $u=\cos(x)$, then $x=\cos^{-1}(u)$ (you will need this later)

and substituting in this last equation you get
$-2u^2-u+1=0$
which has roots $u_1=-1$ and $u_2=1/2$

Then $x_1=\cos^{-1}(-1)=\pi$ and $x_2=\cos^{-1}(1/2)=\pi/3$ or $5\pi/3\$

Those are the 3 solutions

3. thank you guys SO much!