The trigonometric function \[f(x) = \sin x(1 + \cos x)\] has maximum value at x =
${\text{A }}{\text{. 0}}$
${\text{B }}{\text{. }}\pi $
${\text{C }}{\text{. }}\dfrac{\pi }{3}$
${\text{D }}{\text{. 1}}$

Hint – Multiply sin x with the terms inside the bracket and multiply and divide by 2 in the equation, then by using the formula, $2\sin a\cos a = \sin 2a$ and then differentiate it to find the maximum value.

Complete step-by-step solution -
We have been given the function, \[f(x) = \sin x(1 + \cos x)\].
Now multiplying sin x and (1 + cos x), we get-
\[f(x) = \sin x + \sin x\cos x\]
Multiplying and dividing by 2 –
\[f(x) = \sin x + \dfrac{2}{2}\sin x\cos x\]
Now using the trigonometric formula, $2\sin x\cos x = \sin 2x$ , we get-
\[f(x) = \sin x + \dfrac{1}{2}\sin 2x\]
On differentiating w.r.t x, and putting it equal to zero to find the maximum-
\[f'(x) = \cos x + \dfrac{1}{2}.2\cos 2x = \cos x + \cos 2x\]
Now, we know $\cos 2x = 2{\cos ^2}x - 1$ , we get-
\[
  f'(x) = \cos x + \cos 2x \\
   \Rightarrow f'(x) = \cos x + 2{\cos ^2}x - 1 \\
   \Rightarrow f'(x) = 2{\cos ^2}x + \cos x - 1 \\
 \]
To find maxima we put f’(x) = 0, we get-
$2{\cos ^2}x + \cos x - 1 = 0$
Adding and subtracting cos x in the given equation-
$
  2{\cos ^2}x + \cos x + \cos x - \cos x - 1 = 0 \\
   \Rightarrow 2{\cos ^2}x + 2\cos x - \cos x - 1 = 0 \\
   \Rightarrow 2\cos x(\cos x + 1) - 1(\cos x + 1) = 0 \\
   \Rightarrow (2\cos x - 1)(\cos x + 1) = 0 \\
 $
Equating each term to zero we get-
$
  2\cos x - 1 = 0 \\
  \cos x = \dfrac{1}{2} \\
$
And $
  \cos x + 1 = 0 \\
  \cos x = - 1 \\
 $
So, the value of x at $\cos x = \dfrac{1}{2}$ is-
\[ \cos x = \dfrac{1}{2} \\
  \because \cos \dfrac{\pi }{3} = \dfrac{1}{2} \\
   \Rightarrow \cos x = \cos \dfrac{\pi }{3} \\
   \Rightarrow x = \dfrac{\pi }{3} \\ \]
Value of x at $\cos x = - 1$ is –
\[ \cos x = - 1 \\
  \because \cos \pi = - 1 \\
   \Rightarrow \cos x = \cos \pi \\
   \Rightarrow x = \pi \\ \]
$ \Rightarrow x = \dfrac{\pi }{3},\pi $
Again, differentiating f’(x), we get-
 $
  f''(x) = 2.2\cos x.( - \sin x) + ( - \sin x) \\
   = - 4\cos x\sin x - \sin x = - \sin x(4\cos x + 1) \\ $
Now, finding the value of f” (x) at $x = \dfrac{\pi }{3},\pi $
$f''(x){|_{x = \dfrac{\pi }{3}}} = - \sin \dfrac{\pi }{3}(4\cos \dfrac{\pi }{3} + 1) = - \dfrac{{\sqrt 3 }}{2}\left( {4.\dfrac{1}{2} + 1} \right) = - \dfrac{{\sqrt 3 }}{2}\left( {2 + 1} \right) = \dfrac{{ - 3\sqrt 3 }}{2}$
$f''(x){|_{x = \pi }} = - \sin \pi (4\cos \pi + 1) = 0$
So, clearly the function has maximum value at $x = \dfrac{\pi }{3}$ , as f” (x) is negative at $x = \dfrac{\pi }{3}$
Therefore, the value of x at which the given function has maximum is $x = \dfrac{\pi }{3}$ .
Hence, the correct option is ${\text{C }}{\text{. }}\dfrac{\pi }{3}$ .

Note - Whenever such types of questions appear to find the maximum value of the given function we use the concept of minima and maxima where we first, differentiate the given function and then put its derivative equal to zero, the value of variable will be critical point at which we find that the value of second derivative is positive or negative, For positive value we get minima and for negative value we get maxima at the critical point.