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? 
 A: By AM-GM we obtain:
$$\sin\frac{x}{2}+2\sin\frac{x}{4}=2\sin\frac{x}{4}\left(1+\cos\frac{x}{4}\right)=\frac{2}{\sqrt3}\sqrt{\left(3-3\cos\frac{x}{4}\right)\left(1+\cos\frac{x}{4}\right)^3}\leq$$
$$\leq\frac{2}{\sqrt3}\sqrt{\left(\frac{3-3\cos\frac{x}{4}+3\left(1+\cos\frac{x}{4}\right)}{4}\right)^4}=\frac{3\sqrt3}{2}$$
A: you forgot factor of 2 during sum of cosines 
i.e,
$\dfrac{1}{2}\left[cos{\dfrac{x}{2}}+cos{\dfrac{3x}{4}}\right]=cos{\dfrac{3x}{8}}cos{\dfrac{x}{8}}$
make it equal to zero you'will get maxima at $\dfrac{4\pi}{3}$
A: The conclusion that you should have extracted from $f'(x)=\frac12\left(\cos\left(\frac x2\right)+\cos\left(\frac x4\right)\right)$ was that $f'(x)=\cos\left(\frac{3x}8\right)\cos\left(\frac x8\right)$. Therefore, $f'(x)=0\iff x=\frac{4\pi}3$. It turns out that $f''\left(\frac{4\pi}3\right)=-\frac{3\sqrt3}{16}<0$. Therefore, $f$ attains its maximum when $x=\frac{4\pi}3$. Besides, $f\left(\frac{4\pi}3\right)=\frac{3\sqrt3}2$.
A: 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 $f(t) \le \dfrac{3\sqrt{3}}{2}$. 
A: Basically we need to find the maxima of the function in the interval $x \in [0,2\pi]$
$$f (x)=\sin (x/2)+2\sin (x/4)$$
$$f' (x)=\frac {1}{2}\cos(x/2)+\frac {1}{2}\cos (x/4)$$
$$f' (x)=\cos(3x/8)\cos (x/8))$$
Now we need to find the solutions of $f'(x)=0$ in the giiven interval and check the maxima.
$\cos(3x/8)=0\:$ or $\:\cos (x/4)=0$
The only solution to this in the given interval is
$$x=4 \pi/3\:$$
$$f (4 \pi/3)=3\sqrt{3}/2$$
Now we check the end points of the interval just to be sure.
$f (0)=0$ and $f(2 \pi)=2$
So, $f (x)\leq3\sqrt{3}/2$ for all $x \in [0,2\pi]$
A: Hint:
Set $x=8y$  to find $$2\sin2y(1+\cos2y)=8\sin y\cos^3y$$
Now $\sqrt[4]{3\sin^2y\cos^6y}\le\dfrac{3\sin^2y+\cos^2y+\cos^2y+\cos^2y}4=\dfrac34$
Squaring we get  $$16\sqrt3\sin y\cos^3y\le9$$
The equality occurs if $3\sin^2y=\cos^2y\iff\dfrac{\sin^2y}1=\dfrac{\cos^2y}3=\dfrac1{1+3}$
