If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$. Why is my solution wrong? The problem says:

If $450^\circ<\alpha<540^\circ$ and $\cot\alpha=-\frac{7}{24},$ calculate $\cos\frac{\alpha}{2}$

I solved it in the following way: $$\begin{align} -\frac{7}{24}&=-\sqrt{\frac{1+\cos2\alpha}{1-\cos2\alpha}}\\ \frac{49}{576}&=\frac{1+\cos2\alpha}{1-\cos2\alpha}\\ 625\cos2\alpha&=527\\ 2\cos^2\alpha-1&=\frac{527}{625}\\ \cos\alpha&=-\frac{24}{25}, \end{align}$$ therefore, $$\begin{align} \cos\frac{\alpha}{2}&=\sqrt{\frac{1-\frac{24}{25}}{2}}\\ &=\sqrt{\frac{1}{50}}\\ &=\frac{\sqrt{2}}{10}. \end{align}$$ But there is not such an answer:

A) $0.6$
B) $\frac{4}{5}$
C) $-\frac{4}{5}$
D) $-0.6$
E) $0.96$

I have checked the evaluating process several times. While I believe that my answer is correct and there is a mistake in the choices, I want to hear from you.

$$\cot x =\frac{-7}{24}=\frac{\cos x}{\sin x}$$
$$\frac{49}{576}=\frac{\cos^2 x}{1-\cos^2 x}$$
It gives $$\cos x=\frac{-7}{25}$$
And 
$$\cos \frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\frac{3}{5}=0.6 $$ 
Hence (A) is correct 
EDIT:
After concerning with experts , I concluded that $\cos \frac{x}{2} $ is negative and hence (D) is correct
