The period of the function `f(x)=4sin^4((4x-3pi)/(6pi^2))+2cos((4x-3pi)/(3pi^2))` is

Options

A. `(3pi^(2))/(4)`
B. `(3pi^(3))/(4)`
C. `(4pi^(2))/(3)`
D. `(4pi^(3))/(3)`

Answer

To find the period of the function \( f(x) = 4\sin^4\left(\frac{4x - 3\pi}{6\pi^2}\right) + 2\cos\left(\frac{4x - 3\pi}{3\pi^2}\right) \), we will follow these steps:

### Step 1: Identify the inner functions
The function consists of two parts: \( \sin^4\left(\frac{4x - 3\pi}{6\pi^2}\right) \) and \( \cos\left(\frac{4x - 3\pi}{3\pi^2}\right) \). We need to find the periods of these inner functions.

### Step 2: Find the period of the sine function
The argument of the sine function is \( \frac{4x - 3\pi}{6\pi^2} \). To find the period, we can express it in the form \( \sin(kx) \), where \( k \) is the coefficient of \( x \).

1. Rewrite the argument:
\[ \theta = \frac{4x - 3\pi}{6\pi^2} \]
This can be rearranged to:
\[ \theta = \frac{4}{6\pi^2}x - \frac{3\pi}{6\pi^2} \]
Thus, the coefficient of \( x \) is \( k = \frac{4}{6\pi^2} = \frac{2}{3\pi^2} \).

2. The period of \( \sin(kx) \) is given by:
\[ T = \frac{2\pi}{k} = \frac{2\pi}{\frac{2}{3\pi^2}} = 2\pi \cdot \frac{3\pi^2}{2} = 3\pi^3 \]

### Step 3: Find the period of the cosine function
Now, we consider the cosine function \( \cos\left(\frac{4x - 3\pi}{3\pi^2}\right) \).

1. Rewrite the argument:
\[ \phi = \frac{4x - 3\pi}{3\pi^2} \]
This can be rearranged to:
\[ \phi = \frac{4}{3\pi^2}x - \frac{3\pi}{3\pi^2} \]
Thus, the coefficient of \( x \) is \( k = \frac{4}{3\pi^2} \).

2. The period of \( \cos(kx) \) is given by:
\[ T = \frac{2\pi}{k} = \frac{2\pi}{\frac{4}{3\pi^2}} = 2\pi \cdot \frac{3\pi^2}{4} = \frac{3\pi^3}{2} \]

### Step 4: Determine the overall period
The overall period of the function \( f(x) \) will be the least common multiple (LCM) of the two periods we found:

1. The period of the sine part is \( 3\pi^3 \).
2. The period of the cosine part is \( \frac{3\pi^3}{2} \).

To find the LCM of \( 3\pi^3 \) and \( \frac{3\pi^3}{2} \):
- The LCM is \( 3\pi^3 \) because it is the larger of the two periods.

### Final Answer
Thus, the period of the function \( f(x) \) is:
\[ \boxed{3\pi^3} \]