Sine function whose period is 6 is
A) $\sin \dfrac{{2\pi x}}{3}$
B) $\sin \dfrac{{\pi x}}{3}$
C) $\sin \dfrac{{\pi x}}{6}$
D) $\sin \dfrac{{3\pi x}}{2}$

Hint:
We can assume that $f\left( x \right) = \sin kx$ is the required function. Then we can find its period in terms of k. Then we can equate the period to 6 and solve for k. We can substitute the value of k in the function to get the required function.

Complete step by step solution:
Let us assume $f\left( x \right)$ to be the required function. As it is a sin function, we can write it as,
 $f\left( x \right) = \sin kx$
We know that all trigonometric functions have a period of $2\pi $
So, the given function has a period of $\dfrac{{2\pi }}{k}$
We are given that the function has a period of 6. So, we can equate the period of the function to 6 and find the value of k.
 $ \Rightarrow \dfrac{{2\pi }}{k} = 6$
On cross multiplying, we get
 $ \Rightarrow k = \dfrac{{2\pi }}{6}$
On cancelling the common terms, we get
 $ \Rightarrow k = \dfrac{\pi }{3}$
On substituting the value of k in the function, we get
 $f\left( x \right) = \sin \dfrac{{\pi x}}{3}$

Therefore, the required function with period 6 is $\sin \dfrac{{\pi x}}{3}$
So, the correct answer is option B.

Note:
Alternatively, we are given that the period of the function is 6.
 $ \Rightarrow f\left( x \right) = f\left( {6 + x} \right)$
Now we can check through the options.
So, option A will become,
 $f\left( {6 + x} \right) = \sin \dfrac{{2\pi \left( {6 + x} \right)}}{3}$
On expanding the brackets, we get
 $ \Rightarrow f\left( {6 + x} \right) = \sin \left( {\dfrac{{2\pi \times 6}}{3} + \dfrac{{2\pi x}}{3}} \right)$
On simplification we have,
 $ \Rightarrow f\left( {6 + x} \right) = \sin \left( {4\pi + \dfrac{{2\pi x}}{3}} \right)$
This cannot be the answer as the function will have period 6 after it completes 2 cycles.
Now check option B.
 \[f\left( x \right) = \sin \left( {\dfrac{{\pi x}}{3}} \right)\]
 $ \Rightarrow f\left( {6 + x} \right) = \sin \dfrac{{\pi \left( {6 + x} \right)}}{3}$
On expanding the brackets, we get
 $ \Rightarrow f\left( {6 + x} \right) = \sin \left( {\dfrac{{\pi \times 6}}{3} + \dfrac{{\pi x}}{3}} \right)$
On simplification we have,
 $ \Rightarrow f\left( {6 + x} \right) = \sin \left( {2\pi + \dfrac{{\pi x}}{3}} \right)$
We know that the trigonometric functions have a period of $2\pi $
 $ \Rightarrow f\left( {6 + x} \right) = \sin \left( {\dfrac{{\pi x}}{3}} \right)$
 $ \Rightarrow f\left( {6 + x} \right) = f\left( x \right)$
Now check option C.
 \[f\left( x \right) = \sin \left( {\dfrac{{\pi x}}{6}} \right)\]
 $ \Rightarrow f\left( {6 + x} \right) = \sin \dfrac{{\pi \left( {6 + x} \right)}}{6}$
On expanding the brackets, we get
 $ \Rightarrow f\left( {6 + x} \right) = \sin \left( {\dfrac{{\pi \times 6}}{6} + \dfrac{{\pi x}}{6}} \right)$
On simplification we get
 $ \Rightarrow f\left( {6 + x} \right) = \sin \left( {\pi + \dfrac{{\pi x}}{6}} \right)$
But the trigonometric functions have a period of $2\pi $
 $ \Rightarrow \sin \left( {\pi + \dfrac{{\pi x}}{6}} \right) \ne \sin \left( {\dfrac{{\pi x}}{6}} \right)$
 $ \Rightarrow f\left( {6 + x} \right) \ne f\left( x \right)$
So, option C is not a solution.
Now we can consider option D.
 \[f\left( x \right) = \sin \left( {\dfrac{{3\pi x}}{2}} \right)\]
 $ \Rightarrow f\left( {6 + x} \right) = \sin \dfrac{{3\pi \left( {6 + x} \right)}}{2}$
On expanding the brackets, we get
 \[ \Rightarrow f\left( {6 + x} \right) = \sin \left( {\dfrac{{3\pi \times 6}}{2} + \dfrac{{3\pi x}}{2}} \right)\]
On simplification we get,
 $ \Rightarrow f\left( {6 + x} \right) = \sin \left( {9\pi + \dfrac{{\pi x}}{6}} \right)$
But the trigonometric functions have a period of $2\pi $
 $ \Rightarrow \sin \left( {9\pi + \dfrac{{2\pi x}}{3}} \right) \ne \sin \left( {\dfrac{{2\pi x}}{3}} \right)$
 $ \Rightarrow f\left( {6 + x} \right) \ne f\left( x \right)$
So, option D is not a solution.