trigonometry - 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?

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Here is one thing I tried.

Am I completely on the wrong track here?

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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:

$x = \dfrac{\pi}{2}, \dfrac{\pi}{4}, \dfrac{3\pi}{2}, \dfrac{7\pi}{4}$

Answer

Factor out a $\cos(x)$ from your original expression to get:

$$0 = \sqrt{2} \cos^2(x) - \cos(x) = \cos(x)(\sqrt{2}\cos(x)-1)$$

From here, you know $\cos(x)=0$ or $\sqrt{2}\cos(x)-1 = 0$. Do you see where to go from here?