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.

If $x$ is in the third quadrant, its symmetric w.r.t. the $x$-axis is in the second quadrant, and has the same cosine. This symmetric point has curvilinear abscissa $6\pi - x$, as you've found. To determine the value which is represented by the same point on the trigonometric circle, let's just make some elementary manipulations on inequalities.
If $x=10$ˆ, we have $3\pi < x <\dfrac{7\pi}2$, so
\begin{align}
&&\frac{5\pi}2 &<6\pi -x< 3\pi  \\[1ex]
&\llap{\text{whence}}\qquad& \frac \pi 2 &<(6\pi-x)-2\pi=4\pi -x <\pi&&\qquad
\end{align}
The solution follows.
