` 3sin^(2)x-7sin x +2=0, x in [0,(pi)/(2)] and f_(n)(theta)=sin^(n) theta + cos^(n) theta ` .
On the basis of above information, the value of `f_(4)(x) ` is:

Options

A. (a) `(97)/(81)`
B. (b) `(57)/(81)`
C. (c) `(65)/(81)`
D. (d) `(73)/(81)`

Answer

To solve the equation \(3\sin^2 x - 7\sin x + 2 = 0\) for \(x\) in the interval \([0, \frac{\pi}{2}]\) and to find the value of \(f_4(x) = \sin^4 x + \cos^4 x\), we will follow these steps:

### Step 1: Solve the quadratic equation
We start with the equation:
\[ 3\sin^2 x - 7\sin x + 2 = 0 \]
Let \(y = \sin x\). The equation becomes:
\[ 3y^2 - 7y + 2 = 0 \]
We can use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -7\), and \(c = 2\).

### Step 2: Calculate the discriminant
First, we calculate the discriminant:
\[ D = b^2 - 4ac = (-7)^2 - 4 \cdot 3 \cdot 2 = 49 - 24 = 25 \]

### Step 3: Find the roots
Now we can find the roots:
\[ y = \frac{7 \pm \sqrt{25}}{2 \cdot 3} = \frac{7 \pm 5}{6} \]
Calculating the two possible values:
1. \(y_1 = \frac{12}{6} = 2\)
2. \(y_2 = \frac{2}{6} = \frac{1}{3}\)

### Step 4: Determine valid solutions
Since \(y = \sin x\) must be in the range \([0, 1]\), we discard \(y_1 = 2\) as it is not valid. Thus, we have:
\[ \sin x = \frac{1}{3} \]

### Step 5: Calculate \(\cos^2 x\)
Using the identity \(\sin^2 x + \cos^2 x = 1\):
\[ \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]

### Step 6: Calculate \(f_4(x)\)
Now we calculate \(f_4(x) = \sin^4 x + \cos^4 x\):
\[ f_4(x) = \left(\sin^2 x\right)^2 + \left(\cos^2 x\right)^2 = \left(\frac{1}{9}\right)^2 + \left(\frac{8}{9}\right)^2 \]
Calculating each term:
\[ \left(\frac{1}{9}\right)^2 = \frac{1}{81} \]
\[ \left(\frac{8}{9}\right)^2 = \frac{64}{81} \]
Adding these together:
\[ f_4(x) = \frac{1}{81} + \frac{64}{81} = \frac{65}{81} \]

### Final Answer
Thus, the value of \(f_4(x)\) is:
\[ \boxed{\frac{65}{81}} \]