maximum of the function $S(\phi)= \pi(\sin\phi)^2+\pi\sin\phi\sqrt{2+2\cos\phi}$ In a problem I have the function:
$$S(\phi)= \pi(\sin\phi)^2+\pi\sin\phi \sqrt{2+2\cos\phi}$$
in which $\phi \in [0,\pi]$.
I have to find the point of maximum and calculate the maximum value of the function.
$\frac{dS}{d \phi}= 2 \pi*sin(\phi) cos(\phi)+\frac{\pi*((3cos(\phi))^2+2cos(\phi)-1))}{\sqrt{2+2cos(\phi)}}$
Putting this derivative equal to $0$ we have the equation:
$-8 (cos\phi)^5-17 (cos\phi)^4-4(cos\phi)^3+(cos\phi)^2+4cos\phi-1=0\Rightarrow  (cos\phi+1)^3*(-8(cos\phi)^2+7cos\phi-1)=0$
The maximum of this function is for $\cos \varphi = \frac{7+ \sqrt{17}}{16} $
Then $$\sin\varphi=\frac{\sqrt{190-14 \sqrt{17}}}{16}$$
Now I'm trying to find the maximum value of the function but I'm lost in the calculation and I don't know what is the mistake.
The result should be  $$S_{\max}= \frac{\pi(107+51 \sqrt{17})}{128}$$

Note
\begin{align}
S(\phi) &= \pi \sin^2\phi+\pi\sin\phi \sqrt{2+2\cos \phi}\\
&= \pi \sin^2\phi+2\pi\sin\phi \cos \frac\phi2\\
&= 4\pi\left(\sin\frac\phi2 + \sin^2\frac\phi2-\sin^3\frac\phi2-\sin^4\frac\phi2\right)\tag1
\end{align}
Set $S’(\phi)= 0$ to get
\begin{align}
&2\pi \cos\frac\phi2\left( 
4\sin^3\frac\phi2 +3 \sin^2\frac\phi2 - 2\sin\frac\phi2-1\right)\\
= &2\pi \cos\frac\phi2 \left(\sin\frac\phi2+1\right)\left( 4\sin^2\frac\phi2- \sin\frac\phi2-1\right)=0
\end{align}
and the valid solution $\sin\frac\phi2=\frac{1+\sqrt{17}}8$. Then, plug it into (1) to obtain the maximum value
$$S_{max}=\frac{107+51\sqrt{17}}{128}\pi$$
