Show that $\sin (x/2)+2\sin (x/4)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$ 
Given that  $$f (x)=\sin (x/2)+2\sin (x/4)$$
  Show that $f (x)\leq3\sqrt{3}/2$   for all $x \in [0,2\pi]$.

My try 
$f (x)=\sin (x/2)+2\sin (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$
$f' (x)=\frac {1}{2}(\cos(x/2)+\cos (x/4))$
$f' (x)=\cos(3x/4)\cos (x/4))$
$f'(x)=0$   $\implies$  $\cos(3x/4)\cos (x/4))=0$
$\cos(3x/4)=0\:$   or $\:\cos (x/4)=0$
$3x/4= \pi/2\:$  and so $\:x= 2\pi/3$      or $ x=2\pi$
$x/4= \pi/2\:$  and so $\:x= 2\pi $
Is my work ok ? What should I do now? 

Put $\theta = \dfrac{x}{4} \implies \text{ We prove :} f(\theta) = \sin (2\theta)+2\sin(\theta) \le \dfrac{3\sqrt{3}}{2},  \theta \in [0,\pi/2]$ . Put $t = \sin(\theta)\implies t \in [0,1], f(t) = 2t\sqrt{1-t^2}+ 2t\implies f'(t)=2\sqrt{1-t^2}-\dfrac{2t^2}{\sqrt{1-t^2}}+2= 0 \iff t = 0, \dfrac{\sqrt{3}}{2}$ . We have $f(0) = 0, f(1) = 2, f(\frac{\sqrt{3}}{2}) = \dfrac{3\sqrt{3}}{2}$, and this is the largest value and it means $f(t) \le \dfrac{3\sqrt{3}}{2}$. 
