If the middle term in the expansion of `((1)/(x)+x sinx)^(10)` is equal to `7(7)/(8)`, then the number of values of x in `[0, 2pi]` is equal to

Answer

To solve the problem, we need to find the middle term in the expansion of \(\left(\frac{1}{x} + x \sin x\right)^{10}\) and set it equal to \(\frac{7}{8}\). We will then determine the values of \(x\) in the interval \([0, 2\pi]\).

### Step 1: Identify the middle term in the expansion
For the expansion of \((a + b)^n\), the middle term when \(n\) is even is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
where \(r = \frac{n}{2}\).

Here, \(n = 10\), so \(r = \frac{10}{2} = 5\). The middle term \(T_6\) is:
\[ T_6 = \binom{10}{5} \left(\frac{1}{x}\right)^{10-5} \left(x \sin x\right)^5 \]

### Step 2: Calculate the middle term
Substituting the values into the formula:
\[ T_6 = \binom{10}{5} \left(\frac{1}{x}\right)^{5} \left(x \sin x\right)^{5} \]
\[ = \binom{10}{5} \cdot \frac{(x \sin x)^5}{x^5} \]
\[ = \binom{10}{5} \cdot \sin^5 x \]

### Step 3: Calculate \(\binom{10}{5}\)
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 4: Set the middle term equal to \(\frac{7}{8}\)
Now we set the middle term equal to \(\frac{7}{8}\):
\[ 252 \sin^5 x = \frac{7}{8} \]

### Step 5: Solve for \(\sin^5 x\)
Rearranging gives:
\[ \sin^5 x = \frac{7}{8 \times 252} = \frac{7}{2016} \]

### Step 6: Solve for \(\sin x\)
Taking the fifth root:
\[ \sin x = \left(\frac{7}{2016}\right)^{\frac{1}{5}} \]

### Step 7: Finding the values of \(x\)
The sine function is positive in the first and second quadrants. Therefore, we have:
1. \(x = \arcsin\left(\left(\frac{7}{2016}\right)^{\frac{1}{5}}\right)\)
2. \(x = \pi - \arcsin\left(\left(\frac{7}{2016}\right)^{\frac{1}{5}}\right)\)

### Step 8: Count the values in \([0, 2\pi]\)
In the interval \([0, 2\pi]\), the sine function will yield two solutions:
1. One in the first quadrant
2. One in the second quadrant

Thus, the total number of values of \(x\) in the interval \([0, 2\pi]\) is **2**.

### Final Answer
The number of values of \(x\) in \([0, 2\pi]\) is **2**.