# The maximum value of $f(x) = \sin x(1 + \cos x)$A. $3\dfrac{{\sqrt 3 }}{4}$B. $3\dfrac{{\sqrt 3 }}{2}$C. $3\sqrt 3$D. $\sqrt 3$

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Hint:We can use various trigonometric identities such as ${\cos ^2}x - {\sin ^2}x = \cos 2x$ and $\cos 2x = 2{\cos ^2}x - 1$ . Try to factorize for the function as much as possible to get all the possibilities. Check all the possible values of x in the given trigonometry functions.

$f(x) = \sin x(1 + \cos x)$
On Taking the derivative with respect to $x$ on both the sides we get
$f'(x) = \sin x( - \sin x) + (1 + \cos x)\cos x$
Now On multiplying the respective terms we get
$f'(x) = - {\sin ^2}x + \cos x + {\cos ^2}x$
Now Using the identity ${\cos ^2}x - {\sin ^2}x = \cos 2x$ we get
$f'(x) = \cos 2x + \cos x$
Now Using another value for $\cos 2x = 2{\cos ^2}x - 1$ we get
$f'(x) = 2{\cos ^2}x - 1 + \cos x$
Now Splitting the value of $\cos x$ as $2\cos x - \cos x$ we get
$f'(x) = 2{\cos ^2}x + 2\cos x - \cos x - 1$
Now On arranging the terms we get
$f'(x) = 2\cos x(\cos x + 1) - 1(\cos x + 1)$
Now On factorization we get
$f'(x) = (2\cos x - 1)(\cos x + 1)$
Now, let $f'(x) = 0$ to find all the critical points of the function
Then we get
$(2\cos x - 1)(\cos x + 1) = 0$
Then either $\cos x = \dfrac{1}{2}$ or $\cos x = - 1$
Now we know that The maximum of these two values is $\cos x = \dfrac{1}{2}$ which occurs at $x = \dfrac{\pi }{3}$.
Therefore the maximum value of $f(x)$ is at $x = \dfrac{\pi }{3}$.
Hence, $f\left( {\dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right)\left( {1 + \cos \left( {\dfrac{\pi }{3}} \right)} \right)$
$f\left( {\dfrac{\pi }{3}} \right)= \left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {1 + \dfrac{1}{2}} \right) \\ \therefore f\left( {\dfrac{\pi }{3}} \right)= 3\dfrac{{\sqrt 3 }}{4}$
If you find the above solution to be a difficult one you can follow the alternate answer which is as follows.

Therefore option A is the correct answer.

Note: Alternate solution of the given question :
$f(x) = \sin x(1 + \cos x)$ which can be rewritten as $f(x) = \sin x\left( {\dfrac{1}{2}\sin 2x} \right)$
On Taking the derivative with respect to x on both the sides we get
$f'(x) = \cos x + \cos 2x$
Using the CD formula we get
$f'(x) = 2\cos \dfrac{{3x}}{2}\cos \dfrac{x}{2}$
Now, let $f'(x) = 0$
Then we get $2\cos \dfrac{{3x}}{2}\cos \dfrac{x}{2}= 0$
Therefore either $\cos \dfrac{x}{2} = 0$ or $\cos \dfrac{{3x}}{2} = 0$
Therefore we get $x = \pi$ or $x = \dfrac{\pi }{3}$
$f\left( {\dfrac{\pi }{3}} \right) = \sin \left( {\dfrac{\pi }{3}} \right)\left( {1 + \cos \left( {\dfrac{\pi }{3}} \right)} \right) \\ \Rightarrow f\left( {\dfrac{\pi }{3}} \right)= \left( {\dfrac{{\sqrt 3 }}{2}} \right)\left( {1 + \dfrac{1}{2}} \right) \\ \Rightarrow f\left( {\dfrac{\pi }{3}} \right)= 3\dfrac{{\sqrt 3 }}{4}$
$\Rightarrow f\left( \pi \right) = \sin \pi \left( {1 + \cos \left( \pi \right)} \right) = 0$
Therefore it is clearly visible that the maximum value of $f(x)$ is at $x = \dfrac{\pi }{3}$.