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.
 A: If you put $t=\tan \frac {\theta}2$ you obtain $$4\cdot\sin \theta+4\cdot \sin \theta\cdot\cos\theta=4\left(\frac {2t}{1+t^2}\right)\left(1+\frac{1-t^2}{1+t^2}\right)=5$$
Multiply though by $(1+t^2)^2$ to obtain $$16t=5(1+t^2)^2$$From which it is clear that any solution has $t$ positive (rhs is positive), and a quick sketch graph shows there will be two solutions.
$t=0, 16t=0, 5(1+t^2)^2=5\gt 0$
$t=.5, 16t=8, 5(1+t^2)^2=\frac {125}{16}\lt 8$
$t=1, 16t=16, 5(1+t^2)^2=20\gt 16$
So there is one solution for $t$ in $(0,0.5)$ and another in $(0.5,1)$
The equation can be rewritten as a quartic $$5t^4+10t^2-16t+5=0$$
A: Analytically this is almost impossible. Let me explain why:
I started writing $\sin(2x) = 2\cos(x)\sin(x)$ which gave the following equation after factorising :
$$\sin(x)\big(1+\cos(x)\big) = \frac54$$
Now, letting $\cos(x) = \sqrt{1-\sin^2(x)}$ and after taking out the square root, I obtained :
$$\bigg(\frac5{4\sin(x)} - 1\bigg)^2 = 1-\sin^2(x)$$
Expand out and substituting $X = \sin(x)$ you obtain the following equation to solve :
$$X^4 - \frac{5X}{2} + \frac{25}{16} = 0$$
A quartic equation. It is well known that they have a general solution but it is so complicated I'd rather Wolfram do it for me.
Solutions are the inverse sine of them.
From my point of view, there is no easy way to write down the solution, perhaps numerical solution is better in this case. Unless I'm missing an obvious trick...
A: \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}
