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

\(14cosec^2x – 2sin^2x = 21 – 4cos^2x\)
\(\frac {14}{sin^2⁡x}−2sin^2⁡ x = 21−4(1−sin^2⁡x)\)
Let \(sin^2x = t\)
\(⇒ 14 – 2t^2 = 21t – 4t + 4t^2\)
\(⇒ 6t^2 + 17t – 14 = 0\)
\(⇒ 6t^2 + 21t – 4t – 14 = 0\)
\(⇒ 3t(2t + 7) – 2(2t + 7) = 0\)
\(⇒ (2t + 7) (3t– 2) = 0\)
\(⇒t = \frac 23\ or\ -\frac {7}{2}\)
\(⇒ sin^2⁡x=\frac 23\ or −\frac 72\) (not cosider)
\(⇒ sin⁡\ x=±\sqrt {\frac 23}\)

Therefore, \(sin⁡\ x=±\sqrt {\frac 23}\) has \(4\) solutions in the interval \((\frac \pi 4, \frac {7\pi }{4})\).