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

$14cosec^2x – 2sin^2x = 21 – 4cos^2x$
$\frac {14}{sin^2⁡x}−2sin^2⁡ x = 21−4(1−sin^2⁡x)$
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 $(\frac \pi 4, \frac {7\pi }{4})$.