Find the maximum value of $3\sin{\theta \over 2} (1+\cos{\theta} )$ 
Find the maximum value of $3(\sin{\theta \over 2}) (1+\cos{\theta} )$.

I tried A.M-GM. but the equality doesn't hold for any value . Please help me!
 A: Building upon OP's idea of AM-GM, let $\,t = \sin \dfrac{\theta}{2}\,$, then $\,3\sin{\dfrac{\theta}{2}} (1+\cos{\theta} ) = 6t(1-t^2)\,$, and the square of the expression is:
$$\require{cancel}
36 t^2(1-t^2)^2 = 18 \cdot (2t^2)(1-t^2)(1-t^2) \le 18 \cdot \left(\frac{\cancel{(2t^2)}+(1-\cancel{t^2})+(1-\cancel{t^2})}{3}\right)^3 = \frac{16}{3}
$$
Therefore the maximum is $\,\sqrt{\dfrac{16}{3}}=\dfrac{4}{\sqrt{3}}\,$, attained at $\,2t^2=1-t^2\,$ i.e. $\,t = \dfrac{1}{\sqrt{3}}=\sin \dfrac{\theta}{2}\,$.
A: This is very similar to dxiv's solution but does not use $AM-GM$.
Using the double angle formula
$$f=3 \sin \left(\frac{\theta }{2}\right) (1+\cos (\theta ))=3 \sin \left(\frac{\theta }{2}\right)\left(1+1-2\sin^2 \left(\frac{\theta }{2}\right)\right)=6 \sin \left(\frac{\theta }{2}\right) \left(1-\sin ^2\left(\frac{\theta
   }{2}\right)\right)$$
Let $t=\sin \left(\frac{\theta }{2}\right)$ to make 
$$f=6t(1-t^2)=6t-6t^3$$
$$f'=6-18t^2$$
$$f''=-36t$$
So, the first derivative cancels for $t=\pm\frac{1}{\sqrt{3}}$ and since you look for the maximum, you need $f'' <0$ then $t=\frac{1}{\sqrt{3}}$ is the solution. 
So, the maximum value is $\frac{4}{\sqrt{3}}$ and it will be reached at $\theta=2 \sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ (the value of which is not asked).
A: The following Mathematica code
Maximize[3*Sin[x/2]*(1 + Cos[x]), x]

gives
the maximum
$$(x^*,y^*) = \left(4 \tan
   ^{-1}\left(\sqrt{5-2 \sqrt{6}}\right),3 \sin \left(2 \tan ^{-1}\left(\sqrt{5-2 \sqrt{6}}\right)\right) \left(1+\cos
   \left(4 \tan ^{-1}\left(\sqrt{5-2 \sqrt{6}}\right)\right)\right)\right)$$
or
$$(x^*,y^*) \approx (1.23096,2.3094)$$
