What's wrong in my approach in evaluating $\cos^{-1}(\cos\ 10)$ The Question 
$\large\cos^{-1}(\cos 10)$

My Approach
$\large\cos^{-1}(\cos\ (10-2\pi))$ As $2\pi$ is the period of $\cos x.$ Now our angle is-

Now we know that for $\large\cos^{-1}(\cos x) = x$, $\large x\ \epsilon\ [0,\pi]$ but currently our $\large x$ is in the 3rd quadrant, we need to get it into 1st or 2nd quadrant. Therefore let us subtract $\large \pi$ from our current angle i.e $\large 10-2\pi$ 
$\large\cos^{-1}(\cos\ -((10-2\pi)-\pi))$ Now you must be wondering why did I put this minus sign, w.k.t $\large \cos x$ is -ve in third quadrant, by subtracting $\large \pi$ we brought it into 1st quadrant and $\large \cos x$ is +ve in 1st quad, so to compensate that I have put a minus sign.

Now its within range so we can use $\large\cos^{-1}(\cos x) = x$. 
So Final answer $\large -((10-2\pi)-\pi)$ = $\large 3\pi - 10$
But this answer is wrong and the correct answer is $\large 4\pi - 10$.
They have used the property $\large \cos\  (\pi - \theta) = \cos\  (\pi + \theta)$ But why my approach is wrong?
Kindly help.

I suppose that $10^c$ means 10 radians. 
$\pi^c=180^o\implies 1^c \sim57^o\implies 10^c\sim 570^o$
$\cos 10 =\cos (10-2\pi)=-\cos(10-3\pi)=\cos (\pi-(10-3\pi))$
$\arccos(\cos 10)=4\pi-10$ 
Your answer turns out to be wrong as $\arccos(\cos x)=x$ when $x\in [0,\pi]$. Note that $3\pi-10\notin [0,\pi]$.
