Explanation

(a)
Given,
$f(x) = \sin x(1 + cosx)$
$\Rightarrow \ \ \ \ \ \ f(x) = \sin x + \frac{1}{2}\sin{2x}$
On differentiating w.r.t. $x,$ we get
$f'(x) = \cos x + \cos{2x}$
Put $f^{'}(x) = 0$
$\cos{x + \cos{2x = 0}}$
$\Rightarrow \ \ \ 2\cos{\left( \frac{3x}{2} \right)\cos{\left( \frac{x}{2} \right) = 0}}$
$\Rightarrow \ \ \ \cos\frac{x}{2} = 0,\ \ \ \ \ \ \cos\frac{3x}{2} = 0$
$\Rightarrow \ \ \ x = \pi,\ x = \frac{\pi}{3}$
Now, $f^{''}(x) = - \sin x - 2\sin{2x}$
At $x = \frac{\pi}{3},\ f^{''}(x) = - \sin{\frac{\pi}{3} - 2\sin\frac{2\pi}{3}}$
$= - \frac{\sqrt{3}}{2} - \sqrt{3} < 0$, maxima
$\therefore$ maximum value
$f\left( \frac{\pi}{3} \right) = \sin\left( \frac{\pi}{3} \right)\left( 1 + \cos\frac{\pi}{3} \right)$
$= \frac{\sqrt{3}}{2}\left( 1 + \frac{1}{2} \right)$
$= \frac{3\sqrt{3}}{4} = \frac{3^{\frac{3}{2}}}{4}$