Yes I already checked, the answer is x=43°, x=403°... would there be two more answers like you said, or were you just talking about the results before checking for extraneous?
You really need be familiar with trig functions of special angles. The ratio $\displaystyle \frac{\sqrt{3}}{2}$ can be found from the 30-60-90 special triangle. In our case,
$\displaystyle cos 30^{o} = \frac{\sqrt{3}}{2}$.
Also know the cartesian plane. Cosine is positive in Quadrants I and IV, so you can find the angle related to 30 degrees in Quadrant IV by subtracting it from 360 degrees. So
$\displaystyle cos 330^{o} = \frac{\sqrt{3}}{2}$
is also true.
It looks like you're already familiar with coterminal angles, so between that and what I have above, you should be able to find the four solutions.
What's missing is an understanding of the symmetry of the cosine function. You can flip it about a vertical axis of $\displaystyle x=n\pi$, and its graph will look the same. $\displaystyle 2\pi$ is the smallest number $\displaystyle y$ such that $\displaystyle \cos(x)=\cos(x+y)$ for all $\displaystyle x\in\mathbb{R}.$ So trying to find solutions based on any smaller periodicity of the cosine function isn't going to work, I'm afraid.
So, you've got one answer you know is right: 43 degrees. Flip that about the $\displaystyle \pi$ axis. What do you get?
I don't know why you are adding 90 degrees.
$\displaystyle cos 30^{o} = \frac{\sqrt{3}}{2}$
So if
$\displaystyle cos (x - 13^{o}) = \frac{\sqrt{3}}{2}$
then 30° = x - 13°, or x = 43°, which you got.
But
$\displaystyle cos 330^{o} = \frac{\sqrt{3}}{2}$ as well.
So what is x in this case?
Maybe you can draw triangles in the 1st and 4th quadrants? sqrt(3) over 2 is a very common trig ratio, so if I were to be grading this problem I probably would accept work like this:
$\displaystyle \begin{aligned}
cos (x - 13^{o}) &= \frac{\sqrt{3}}{2} \\
cos (x - 13^{o}) &= cos 30^{o}\,&\text{OR}\,cos (x - 13^{o}) &= cos 330^{o} \\
x - 13^{o} &= 30^{o}\,&\text{OR}\,x - 13^{o} &= 330^{o} \\
\end{aligned}$
... and so on.
BTW, it's not x = 317. Add 13 degrees, not subtract.
Cosine is positive in the first and fourth quadrants.
Have you learned about reference angles? Take a look at this page from Spark Notes:
SparkNotes: Trigonometry: Trigonometric Functions: Reference Angles
I know that $\displaystyle cos 30^{o} = \frac{\sqrt{3}}{2}$, and that cosine is also positive in the 4th quadrant. To find the angle whose reference angle is 30°, subtract 30° from 360° (this is true for any angle in the 4th quadrant) and you get 330°. So $\displaystyle cos 330^{o} = \frac{\sqrt{3}}{2}$ is also true.