Problem with trigonometric equation I am having trouble solving this equation
$$4\cdot \sin \theta + 2 \cdot \sin 2\theta =5$$
Thank you for your help.

\begin{align}
4\sin \theta+2\sin 2\theta&=5\\
4\sin \theta+2\cdot2\sin \theta\cos\theta&=5\\
4\sin \theta+4\sin \theta\cos\theta&=5\\
4\sin \theta(1+\cos\theta)&=5
\end{align}
Take square both sides.
\begin{align}
4^2\sin^2 \theta(1+\cos\theta)^2&=5^2\\
16(1-\cos^2\theta)(1+\cos\theta)^2&=25
\end{align}
Let $x=\cos\theta$.
\begin{align}
16(1-x^2)(1+x)^2&=25\\
x^4+2x^3-2x-1&=-\frac{25}{16}\\
x^4+2x^3-2x&=-\frac{9}{16}\\
16x^4+32x^3-32x+9&=0
\end{align}
