Maximize $\frac{\sin(2x)}{3-\cos(2x)}$ I want to maximize $\frac{\sin(2x)}{3-\cos(2x)}$. I converted both $\sin$ and $\cos$ to $\tan$ and got answer $$\frac{1}{2\sqrt 2}$$ achieved for example at $x=\operatorname{arccot}(\sqrt2)$. Please give some another approach without using the above conversion formula.
Here is another Approach : 

Since a point $(\cos (2t), \sin (2t))$ is on a circle $x^2 + y^2 = 1$,

we can understand $\frac{\sin (2x)}{3-\cos(2x)} = \frac{0-\sin (2x)}{3-\cos(2x)}$ by slope of tangent line between a point $(3, 0)$ and a circle $x^2+y^2=1$.

After calculations, we can get :$$\max\left\{\frac{\sin (2x)}{3-\cos(2x)} \right\} = \frac{1}{2\sqrt{2}} \\
\min\left\{\frac{\sin (2x)}{3-\cos(2x)} \right\} = -\frac{1}{2\sqrt{2}}$$
Hint

Use https://en.m.wikipedia.org/wiki/Tangent_half-angle_substitution#The_substitution
to replace $\sin2x,\cos2x$ with $$\dfrac{2t}{1+t^2},\dfrac{1-t^2}{1+t^2}$$ respectively where $t=\tan x$ to  find

$\dfrac{\sin2x}{3-\cos2x}=\dfrac{2t}{3(1+t^2)-(1-t^2)}=y$(say)

Method$\#1:$

Rearrange the relationship to form a quadratic equation in $t$

As $t$ is real, the discriminant must be $\ge0$

Method$\#2:$

$$y=\dfrac t{1+2t^2}$$

If $t\ge0,$ $$\dfrac{1+2t^2}t\ge2\sqrt{\dfrac1t\cdot2t}=?$$

If $t&lt;0,$ set $t=-u$