The maximum value of function \[f(x)=\sin x(1+\cos x)\] \[x\in R\] is:
A. $\dfrac{{{\text{3}}^{\dfrac{\text{3}}{\text{2}}}}}{\text{4}}$
B. $\dfrac{{{\text{3}}^{\dfrac{\text{5}}{\text{3}}}}}{\text{4}}$
C. $\dfrac{\text{3}}{2}$
D. $\dfrac{{{\text{3}}^{\dfrac{7}{5}}}}{\text{4}}$

Hint: Firstly multiply the entire function then apply the first derivative test then equate the obtained term to zero to find the values of \[x\].Then apply the second derivative test to find the maximum value and then put the obtained value in the given solution to check which option is correct.

Complete step by step answer:
In our daily life, there are some situations where the greatest and least values of a function are required. The value of a function \[f(x)\] of \[x\] is said to be maximum when it is greater than the value which the function can assume immediately preceding or following. If the value of a function will increase when the value of the independent variable \[x\] will increase. After retaining maximum value its value will start to decrease, then the points at which the functions will start to decrease is called maxima of the function. In a similar way the point at which the value of the function will start to increase, is called minima of the function.
For maxima : At \[x=a\], the value of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] should be negative that is \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}<0\]
For minima : At \[x=a\], the value of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\] should be positive that is \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}>0\]
Now according to the question:
\[f(x)=\sin x(1+\cos x)\]
Now we are multiplying the entire function we will get:
\[f(x)=\sin x+\sin x\cdot \cos x\]
Now differentiating the function \[f(x)\] with respect to \[x\] on both sides we get:
\[\dfrac{d}{dx}f(x)=\dfrac{d}{dx}\sin x+\dfrac{d}{dx}\sin x\cdot \cos x\]
The product rule always says that the derivative of \[fg=fg'+f'g\]
Hence,\[f'(x)=\cos x+\sin x\dfrac{d}{dx}\cos x+\cos x\dfrac{d}{dx}\sin x\]
\[f'(x)=\cos x+\sin x(-\sin x)+\cos x(\cos x)\]
\[f'(x)=\cos x-{{\sin }^{2}}x+{{\cos }^{2}}x\]
Now we know that \[\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x\]
So, \[f'(x)=\cos x+\cos 2x\]
We also know that \[\cos 2x=2{{\cos }^{2}}x-1\], putting the value we will get:
\[f'(x)=\cos x+2{{\cos }^{2}}x-1\]
We may write \[\cos x\] as \[2\cos x-\cos x\]
So, \[f'(x)=2\cos x-\cos x+2{{\cos }^{2}}x-1\]
\[f'(x)=2\cos x+2{{\cos }^{2}}x-\cos x-1\]
Taking \[2\cos x\] and \[-1\] in common we get:
\[f'(x)=2\cos x(\cos x+1)-1(\cos x+1)\]
\[f'(x)=(2\cos x-1)(\cos x+1)\]
Equating each term to zero we get:
\[2\cos x-1=0\]
\[\cos x=\dfrac{1}{2}\]
And \[\cos x+1=0\]
\[\cos x=-1\]
Value of \[x\] at \[\cos x=\dfrac{1}{2}\]
\[\cos \dfrac{\pi }{3}=\cos \dfrac{1}{2}\]
\[\cos x=\cos \dfrac{\pi }{3}\]
\[x=\dfrac{\pi }{3}\]
Value of \[x\] at \[\cos x=-1\]
\[\cos \pi =-1\]
\[\cos x=\cos \pi \]
\[x=\pi \]
\[\Rightarrow x=\dfrac{\pi }{3},\pi \]
 Again differentiating \[f'(x)\] with respect to \[x\]on both sides
\[\dfrac{d}{dx}f'(x)=\dfrac{d}{dx}(2{{\cos }^{2}}x+\cos x-1)\]
\[\dfrac{d}{dx}f'(x)=\dfrac{d}{dx}2{{\cos }^{2}}x+\dfrac{d}{dx}\cos x-\dfrac{d}{dx}1\]
\[f''(x)=-4\cos (x)\sin x-\sin x\]
Putting the values of \[x\] in \[f''(x)\]
\[{{\left. f''(x) \right|}_{x=\dfrac{\pi }{3}}}=-4\cos (\dfrac{\pi }{3})\sin (\dfrac{\pi }{3})-\sin (\dfrac{\pi }{3})\]
   \[=-4\times \dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{3}}{2}\]
\[=-\dfrac{3\sqrt{3}}{2}\]
\[{{\left. f''(x) \right|}_{x=\pi }}=-4\cos (\pi )\sin (\pi )-\sin (\pi )\]
 \[=-4\times (-1)\times 0-0\]
  \[=0\]
Hence it is cleared that the function has maximum value at \[x=\dfrac{\pi }{3}\], as \[f''(x)\] is negative at \[x=\dfrac{\pi }{3}\].
Hence putting the value of \[x=\dfrac{\pi }{3}\] in the given equation \[f(x)=\sin x(1+\cos x)\]
So, \[f(\dfrac{\pi }{3})=\sin \dfrac{\pi }{3}(1+\cos \dfrac{\pi }{3})\]
 \[=\dfrac{\sqrt{3}}{2}(1+\dfrac{1}{2})\]
\[=\dfrac{3\sqrt{3}}{4}\]
 \[=\dfrac{{{3}^{1}}\times {{3}^{\dfrac{1}{2}}}}{4}\]
\[=\dfrac{{{3}^{1+\dfrac{1}{2}}}}{4}\]
 \[=\dfrac{{{3}^{\dfrac{3}{2}}}}{4}\]

So, the correct answer is “Option A”.

Note: You must remember that a function can have various maximum and minimum values and between every two maximum values there exists a minimum value or vice versa and if you want to identify the local maxima or minima of the given function, at first you should try to apply the first derivative test.