By the defintion of a geometric sequence, we have $\cos^2x=\sin x \tan x$. Since $\tan x=\frac{\sin x}{\cos x}$, we can rewrite this as $\cos^3x=\sin^2x$.
The common ratio of the sequence is $\frac{\cos x}{\sin x}$, so we can write
$$a_1= \sin x$$
$$a_2= \cos x$$
$$a_3= \frac{\cos^2x}{\sin x}$$
$$a_4=\frac{\cos^3x}{\sin^2x}=1$$
$$a_5=\frac{\cos x}{\sin x}$$
$$a_6=\frac{\cos^2x}{\sin^2x}$$
$$a_7=\frac{\cos^3x}{\sin^3x}=\frac{1}{\sin x}$$
$$a_8=\frac{\cos x}{\sin^2 x}=\frac{1}{\cos^2 x}$$

Since $\cos^3x=\sin^2x=1-\cos^2x$, we have $\cos^3x+\cos^2x=1 \implies \cos^2x(\cos x+1)=1 \implies \cos x+1=\frac{1}{\cos^2 x}$, which is $a_8$ , making our answer $8 \Rightarrow \boxed{E}$.