Please help in a trigonometry question!

Question:
The number of real solutions to the equation $\left | sec(x)-2\right |=k$, where $-2\pi\le x \le2\pi$

is determined by the value of the constant k.

Find the number of real solutions for all values of k, given that $k\epsilon R$.

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Here is my attempt:

$\frac{1}{cos(x)}-2=\pm k$ for $-2\pi\le x\le 2\pi$

$cos(x)=\frac{1}{2\pm k}$

$x=cos^{-1}(\frac{1}{2\pm k})$ ( $-2\pi\le x\le 2\pi$ ) 

So:

$-2\pi\le cos^{-1}(\frac{1}{2\pm k})\le 2\pi$ implying: $1\le\frac{1}{2\pm k}\le1$.

Therefore, $\frac{1}{2\pm k}=1$, so $k=\pm1$

Now, i subsituited k=1 and k=-1 in the original equation $\left | sec(x)-2\right |=k$ and considering the domain, $-2\pi\le x \le2\pi$

found the solutions seperately for each case (I.e. k=1 and k=-1) Then counted them and they are 7 distinct solutions.

Is this correct? As I do not have a model answer to check. 

Thanks in advance.