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.

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 $$5t^4+10t^2-16t+5=0$$
