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Hint: We will start solving this question by simplifying the given function. We will simplify the exponent in the function and then we will find the period of each term we get after simplification for simplification, we will use the properties of trigonometry.

Complete step-by-step answer:

We will first simplify the term given in the exponent of the function. For simplification, we will use the trigonometric properties of ${\sin ^2}x$.

Now, we know that ${\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}$

Therefore, from the above property, we get

${\sin ^2}\pi x = \dfrac{{1 - \cos 2\pi x}}{2}$

Also, ${\sin ^4}\pi x = {({\sin ^2}\pi x)^2}$

Therefore, using the above property, we get

${\sin ^4}\pi x = {\left( {\dfrac{{1 - \cos 2\pi x}}{2}} \right)^2}$

Using ${(a + b)^2} = {a^2} + {b^2} + 2ab$, we get

${\sin ^4}\pi x = \dfrac{1}{4}(1 + {\cos ^2}2\pi x - 2\cos 2\pi x)$

${\sin ^4}\pi x = \dfrac{3}{8}(3 + \cos 4\pi x - 4\cos 2\pi x)$

Now, the period of $\cos Ax$ is $\dfrac{{2\pi }}{A}$. Therefore, the period of $\cos 2\pi x = \dfrac{{2\pi }}{{2\pi }} = 1$

So, the period of ${\sin ^2}\pi x$ = 1

Similarly, period of $\cos 4\pi x$ = $\dfrac{{2\pi }}{{4\pi }} = \dfrac{1}{2}$ and the period of $\cos 2\pi x = \dfrac{{2\pi }}{{2\pi }} = 1$ .

Therefore, the period of ${\sin ^4}\pi x$ is LCM (1, $\dfrac{1}{2}$)

Now, LCM of two numbers are calculated as LCM of numerator divided by HCF of denominator. Therefore, we get

LCM (1, $\dfrac{1}{2}$) = LCM (1,1)/HCF (1,2) = 1/1 = 1

So, the period of ${\sin ^4}\pi x$ is 1. … (2)

Now, x – [x] = {x}, which is known as a fractional part of a number. The fractional part of the number has a period 1. So, the period of {x} is 1.

Therefore, the period of x – [x] is 1. … (3)

So, from equation (1), (2) and (3), we get

Period of \[{3^{({{\sin }^2}\pi x + x - [x] + {{\sin }^4}\pi x)}}\] is 1.

Note: Whenever we come up with such problems, we will first start by simplifying the given function. We will do this by using properties of trigonometric, logarithmic, exponential, etc. After simplifying the given function, we will find the period of each and every term of the function and after it the period of the function can be found easily. The period should be found correctly by using the correct formula.

Complete step-by-step answer:

We will first simplify the term given in the exponent of the function. For simplification, we will use the trigonometric properties of ${\sin ^2}x$.

Now, we know that ${\sin ^2}x = \dfrac{{1 - \cos 2x}}{2}$

Therefore, from the above property, we get

${\sin ^2}\pi x = \dfrac{{1 - \cos 2\pi x}}{2}$

Also, ${\sin ^4}\pi x = {({\sin ^2}\pi x)^2}$

Therefore, using the above property, we get

${\sin ^4}\pi x = {\left( {\dfrac{{1 - \cos 2\pi x}}{2}} \right)^2}$

Using ${(a + b)^2} = {a^2} + {b^2} + 2ab$, we get

${\sin ^4}\pi x = \dfrac{1}{4}(1 + {\cos ^2}2\pi x - 2\cos 2\pi x)$

${\sin ^4}\pi x = \dfrac{3}{8}(3 + \cos 4\pi x - 4\cos 2\pi x)$

Now, the period of $\cos Ax$ is $\dfrac{{2\pi }}{A}$. Therefore, the period of $\cos 2\pi x = \dfrac{{2\pi }}{{2\pi }} = 1$

So, the period of ${\sin ^2}\pi x$ = 1

Similarly, period of $\cos 4\pi x$ = $\dfrac{{2\pi }}{{4\pi }} = \dfrac{1}{2}$ and the period of $\cos 2\pi x = \dfrac{{2\pi }}{{2\pi }} = 1$ .

Therefore, the period of ${\sin ^4}\pi x$ is LCM (1, $\dfrac{1}{2}$)

Now, LCM of two numbers are calculated as LCM of numerator divided by HCF of denominator. Therefore, we get

LCM (1, $\dfrac{1}{2}$) = LCM (1,1)/HCF (1,2) = 1/1 = 1

So, the period of ${\sin ^4}\pi x$ is 1. … (2)

Now, x – [x] = {x}, which is known as a fractional part of a number. The fractional part of the number has a period 1. So, the period of {x} is 1.

Therefore, the period of x – [x] is 1. … (3)

So, from equation (1), (2) and (3), we get

Period of \[{3^{({{\sin }^2}\pi x + x - [x] + {{\sin }^4}\pi x)}}\] is 1.

Note: Whenever we come up with such problems, we will first start by simplifying the given function. We will do this by using properties of trigonometric, logarithmic, exponential, etc. After simplifying the given function, we will find the period of each and every term of the function and after it the period of the function can be found easily. The period should be found correctly by using the correct formula.

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