Trig identities - stuck solving $\tan^2\theta = -\frac 32 \sec\theta$ Solve the equation on the interval $0\leq \theta < 2\pi$
$$\tan^2 \theta = -\frac{3}{2}\sec \theta $$
Here are the steps I have so far:
Identity: $\tan^2 \theta = \sec^2 \theta -1 $
Substitute:
$$\sec^2 \theta -1  = -\frac{3}{2}\sec \theta $$
$$2\sec^2 \theta -2 = {-3}\sec \theta $$
$$2\sec^2 \theta +3\sec \theta - 2 = 0 $$
Is this factoring correct?:
$$(2\sec\theta+4)(\sec\theta-1) = 0 $$
$$2\sec\theta+4 =0$$
$$2\sec\theta = -4 $$
$$\sec \theta = -2$$
$$(2\pi/3), (4\pi/3) $$
$$\sec\theta - 1 = 0$$
$$\sec\theta = 1 $$
$\sec\theta=1$ would evaluate to $0$, for some reason that is an invalid answer? (according to my assignment)
So is $\ (2\pi/3), (4\pi/3) $ the full answer? 

Set $\cos\theta=t$; then
$$
\tan^2\theta=\frac{1-t^2}{t^2}
\qquad
\sec\theta=\frac{1}{t}
$$
and the equation becomes
$$
\frac{1-t^2}{t^2}=-\frac{3}{2t}
$$
that is,
$$
2-2t^2=-3t
$$
and finally
$$
2t^2-3t-2=0
$$
The roots are $2$ and $-1/2$. So the equation reduces to
$$
\cos\theta=-\frac{1}{2}
$$
