If number of solution and sum of solution of the equation ` 3sin^(2)x-7sinx +2=0, x in [0,2pi]` are respectively N and S and ` f_(n)(theta)=sin^(n)theta +cos^(n) theta`. On the basis of above information , answer the following questions.
If ` alpha ` is solution of equation ` 3sin^(2)x-7sinx+2=0, x in [0,2pi]`, then the value of `f_(4)(alpha)` is

Options

A. `(97)/(81)`
B. `(57)/(81)`
C. `(65)/(81)`
D. 0

Answer

To solve the equation \(3\sin^2 x - 7\sin x + 2 = 0\) for \(x\) in the interval \([0, 2\pi]\), we will follow these steps:

### Step 1: Rewrite the Equation
We start with the equation:
\[ 3\sin^2 x - 7\sin x + 2 = 0 \]
Let \(y = \sin x\). The equation becomes:
\[ 3y^2 - 7y + 2 = 0 \]

### Step 2: Factor the Quadratic Equation
Next, we will factor the quadratic equation:
\[ 3y^2 - 6y - y + 2 = 0 \]
This can be grouped as:
\[ 3y(y - 2) - 1(y - 2) = 0 \]
Factoring out \((y - 2)\):
\[ (3y - 1)(y - 2) = 0 \]

### Step 3: Solve for \(y\)
Setting each factor to zero gives us:
1. \(3y - 1 = 0 \Rightarrow y = \frac{1}{3}\)
2. \(y - 2 = 0 \Rightarrow y = 2\) (not possible since \(\sin x\) must be in the range \([-1, 1]\))

Thus, the only valid solution is:
\[ \sin x = \frac{1}{3} \]

### Step 4: Find the Values of \(x\)
To find \(x\) in the interval \([0, 2\pi]\):
\[ x = \arcsin\left(\frac{1}{3}\right) \quad \text{and} \quad x = \pi - \arcsin\left(\frac{1}{3}\right) \]
This gives us two solutions.

### Step 5: Count the Number of Solutions
The number of solutions \(N\) is:
\[ N = 2 \]

### Step 6: Calculate the Sum of Solutions
The sum of the solutions \(S\) can be calculated as:
\[ S = \arcsin\left(\frac{1}{3}\right) + \left(\pi - \arcsin\left(\frac{1}{3}\right)\right) = \pi \]

### Step 7: Calculate \(f_4(\alpha)\)
Now, we need to find \(f_4(\alpha)\) where:
\[ f_n(\theta) = \sin^n \theta + \cos^n \theta \]
We already have \(\sin x = \frac{1}{3}\). We need to find \(\cos x\):
\[ \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]
Thus,
\[ \cos x = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

Now we can calculate \(f_4(\alpha)\):
\[ f_4(\alpha) = \sin^4 x + \cos^4 x \]
Calculating each term:
\[ \sin^4 x = \left(\frac{1}{3}\right)^4 = \frac{1}{81} \]
\[ \cos^4 x = \left(\frac{2\sqrt{2}}{3}\right)^4 = \frac{16}{81} \]
Thus,
\[ f_4(\alpha) = \frac{1}{81} + \frac{16}{81} = \frac{17}{81} \]

### Final Answer
The value of \(f_4(\alpha)\) is:
\[ \frac{17}{81} \]