The number of values of x in the interval `(pi/4,(7pi)/4)` for which `14"cosec"^(2) x – 2sin^2x = 21 - 4 cos^(2)x` holds, is _______

Answer

To solve the equation \( 14 \csc^2 x - 2 \sin^2 x = 21 - 4 \cos^2 x \) for the number of values of \( x \) in the interval \( \left(\frac{\pi}{4}, \frac{7\pi}{4}\right) \), we can follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We know that \( \csc^2 x = \frac{1}{\sin^2 x} \) and \( \cos^2 x = 1 - \sin^2 x \). Therefore, we can rewrite the equation as:
\[ 14 \frac{1}{\sin^2 x} - 2 \sin^2 x = 21 - 4(1 - \sin^2 x) \]

### Step 2: Simplify the equation
Substituting \( \cos^2 x \) gives:
\[ 14 \frac{1}{\sin^2 x} - 2 \sin^2 x = 21 - 4 + 4 \sin^2 x \]
This simplifies to:
\[ 14 \frac{1}{\sin^2 x} - 2 \sin^2 x = 17 + 4 \sin^2 x \]
Combining like terms, we have:
\[ 14 \frac{1}{\sin^2 x} - 6 \sin^2 x = 17 \]

### Step 3: Multiply through by \( \sin^2 x \)
To eliminate the fraction, multiply the entire equation by \( \sin^2 x \):
\[ 14 - 6 \sin^4 x = 17 \sin^2 x \]
Rearranging gives us:
\[ 6 \sin^4 x + 17 \sin^2 x - 14 = 0 \]

### Step 4: Let \( t = \sin^2 x \)
Let \( t = \sin^2 x \). The equation becomes:
\[ 6t^2 + 17t - 14 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ t = \frac{-17 \pm \sqrt{17^2 - 4 \cdot 6 \cdot (-14)}}{2 \cdot 6} \]
Calculating the discriminant:
\[ 17^2 = 289, \quad 4 \cdot 6 \cdot 14 = 336 \quad \Rightarrow \quad 289 + 336 = 625 \]
Thus,
\[ t = \frac{-17 \pm 25}{12} \]
Calculating the two possible values:
1. \( t = \frac{8}{12} = \frac{2}{3} \)
2. \( t = \frac{-42}{12} = -3.5 \) (not valid since \( t \) must be non-negative)

### Step 6: Find \( \sin x \)
Since \( t = \sin^2 x \), we have:
\[ \sin^2 x = \frac{2}{3} \quad \Rightarrow \quad \sin x = \pm \sqrt{\frac{2}{3}} \]

### Step 7: Determine the values of \( x \)
The solutions for \( \sin x = \sqrt{\frac{2}{3}} \) and \( \sin x = -\sqrt{\frac{2}{3}} \) in the interval \( \left(\frac{\pi}{4}, \frac{7\pi}{4}\right) \) are:
- For \( \sin x = \sqrt{\frac{2}{3}} \):
- \( x = \arcsin\left(\sqrt{\frac{2}{3}}\right) \)
- \( x = \pi - \arcsin\left(\sqrt{\frac{2}{3}}\right) \)

- For \( \sin x = -\sqrt{\frac{2}{3}} \):
- \( x = \frac{3\pi}{2} + \arcsin\left(\sqrt{\frac{2}{3}}\right) \)
- \( x = \frac{3\pi}{2} - \arcsin\left(\sqrt{\frac{2}{3}}\right) \)

### Step 8: Count the solutions
In the interval \( \left(\frac{\pi}{4}, \frac{7\pi}{4}\right) \):
- The two solutions from \( \sin x = \sqrt{\frac{2}{3}} \) are valid.
- The two solutions from \( \sin x = -\sqrt{\frac{2}{3}} \) are also valid.

Thus, the total number of solutions is \( 4 \).

### Final Answer
The number of values of \( x \) in the interval \( \left(\frac{\pi}{4}, \frac{7\pi}{4}\right) \) is **4**.