= Solution
Write $y(\theta)=Y(x)$ with $x=\cos\theta$. By the <chain rule>,
$$
\frac{dy}{d\theta}=-\sin\theta\,Y_x,\qquad\frac{d^2y}{d\theta^2}=-\cos\theta\,Y_x+\sin^2\theta\,Y_{xx}.
$$
The terms involving $Y_x$ cancel in the original <ordinary differential equation>, leaving
$$
\sin\theta\,y_{\theta\theta}-\cos\theta\,y_\theta+2y\sin^3\theta=\sin^3\theta\,(Y_{xx}+2Y)=0.
$$
On each interval where $\sin\theta\ne0$, the transformed <linear ordinary differential equation> is $Y_{xx}+2Y=0$. Its <characteristic equation> has roots $\pm i\sqrt2$, yielding
$$
\boxed{y(\theta)=C_1\cos(\sqrt2\cos\theta)+C_2\sin(\sqrt2\cos\theta).}
$$
Direct substitution shows these expressions also extend smoothly through $\theta=k\pi$ and satisfy the original equation there. The two constants describe the general solution on any interval where the coordinate substitution is invertible.
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