Solve for $x$ in the following trigonometric equation $$\sqrt 2\cos^2 x-\cos x=0$$
Solve for $x$ algebraically, where $x$ is greater than or equal to zero, and less than $2\pi$. Answer must be an exact solution. 
To be honest, I don't know where to start with this one. I know I need to isolate $\cos x$, but I have little idea as to what I need to do to get there. Is subtracting $\cos x$ from both sides the best way to go about this?

Here is one thing I tried. 

Am I completely on the wrong track here?

EDIT:
$\sqrt{2}\cos x - 1 = 0$
$\cos x = \dfrac{1}{\sqrt{2}}$
$\dfrac{1}{\sqrt{2}} = 45^\circ$
$360 - 45 = 315^\circ = \dfrac{7\pi}{4}$
$\cos x = 0$
$x = 0$ at $90^\circ$, or $\dfrac{\pi}{2}$ and $270^\circ$ or $\dfrac{3\pi}{2}$
So:
$x = \dfrac{\pi}{2}, \dfrac{\pi}{4}, \dfrac{3\pi}{2}, \dfrac{7\pi}{4}$

We have: $\sqrt{2}\cos^{2}(x)-\cos(x)=0$
Let's factor out a $\cos(x)$ to get:
$\Rightarrow \cos(x)\big(\sqrt{2}\cos(x)-1\big)=0$
As we have a product that equals to zero, either one of the multiples must equal to zero:
$\Rightarrow \cos(x)=0 \Rightarrow x=\arccos(0)=\dfrac{\pi}{2}$, $\hspace{1 mm}\dfrac{3\pi}{2}$
$\Rightarrow \sqrt{2}\cos(x)-1=0 \Rightarrow x=\arccos\bigg(\dfrac{\sqrt{2}}{2}\bigg)=\dfrac{\pi}{4}$, $\hspace{1 mm}\dfrac{7\pi}{4}$
Therefore, the solutions to the equation are $x=\dfrac{\pi}{2}$, $x=\dfrac{3\pi}{2}$, $x=\dfrac{\pi}{4}$ and $x=\dfrac{7\pi}{4}$.
