Troubling Trigono

The given relation is:

$\sin \theta+\sin^{2} \theta+\sin^{3} \theta=1$

$\therefore \sin \theta+\sin^{3} \theta=\cos^{2} \theta$

Squaring both sides:

$\sin^{2} \theta+\sin^{6} \theta+2\sin^{4} \theta=\cos^{4} \theta$

Replacing $\sin^{2} \theta$ by $1-\cos^{2} \theta$ :

$(1-\cos^{2} \theta)+(1-\cos^{2} \theta)^{3}+2(1-\cos^{2} \theta)^{2}=\cos^{4} \theta$

Now just expanding and rearranging we get:

$\cos^{6} \theta-4\cos^{4} \theta+8\cos^{2} \theta=4$

What I did is that I let sin theta be equal to x, hence we have an equation in the degree 3.. I get the root and since only one root is real (the rest are imaginary), i get the arc sin of the real root.. substituting it to the question, i got 4..