Question

The number of values of x in the interval \((\frac \pi 4, \frac {7\pi }{4})\) for which \(14cosec^2x – 2sin^2x = 21 – 4cos^2x\) holds, is ______.

Answer 1

Correct Answer: 4

To solve the problem, start by simplifying the given equation: \(14\csc^2x - 2\sin^2x = 21 - 4\cos^2x\).

1. Recall the identity: \(\csc^2x = \frac{1}{\sin^2x}\), thus \(14\csc^2x = \frac{14}{\sin^2x}\).
2. Substitute and rearrange the equation:
   \(\frac{14}{\sin^2x} - 2\sin^2x = 21 - 4\cos^2x\).
3. Use the Pythagorean identity \(\cos^2x = 1 - \sin^2x\):
   \(4\cos^2x = 4(1 - \sin^2x) = 4 - 4\sin^2x\), so the equation becomes:
   \(\frac{14}{\sin^2x} - 2\sin^2x = 17 + 4\sin^2x\).
4. Multiply through by \(\sin^2x\) to eliminate the fraction:
   \(14 - 2\sin^4x = 17\sin^2x + 4\sin^4x\).
5. Rearrange terms:
   \(6\sin^4x + 17\sin^2x - 14 = 0\).
6. Let \(y = \sin^2x\), transforming the equation:
   \(6y^2 + 17y - 14 = 0\).
7. Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) where \(a = 6\), \(b = 17\), and \(c = -14\). Compute the discriminant:
   \(b^2 - 4ac = 289 + 336 = 625\).
8. Thus, \(y = \frac{-17 \pm 25}{12}\). Calculate the roots:
   \(y = \frac{8}{12} = \frac{2}{3}\) and \(y = \frac{-42}{12} = -\frac{7}{2}\) (neglect the negative root since \(y = \sin^2x \geq 0\)).
9. So, \(\sin^2x = \frac{2}{3}\). Therefore, \(\sin x = \pm \frac{\sqrt{2}}{\sqrt{3}}\).
10. Determine \(x\) values in the given interval \((\frac{\pi}{4}, \frac{7\pi}{4})\):
    \(\sin^{-1}(\frac{\sqrt{2}}{\sqrt{3}})\) and \(\pi - \sin^{-1}(\frac{\sqrt{2}}{\sqrt{3}})\), which correspond to two angles per quadrant.
11. Given \((\frac{\pi}{4}, \frac{7\pi}{4})\) spans from 1st to 3rd quadrants, count valid \(x\) values. We have two from 1st, one from 2nd, and one from 3rd, totaling to 4 solutions.

Hence, the number of values of \(x\) is 4, which fits the expected range given as 4,4.