Correct Answer - C
Let `(sin 3theta)/(cos 2theta)lt 0`
Case I. `sin theta lt 0` and `cos 2theta gt 0`
`rArr 3theta in (pi, 2pi)` and `2theta in (-(pi)/(2),(-pi)/(2))`
`rArr theta in((pi)/(3),(2pi)/(3))` and `theta in (-(pi)/(4), (pi)/(4))`
Case II. `sin 3theta gt 0` and `cos 2 theta lt 0`
`therefore 3theta in(0, pi)` and `2theta in((pi)/(2),(3pi)/(2))`
`rArr theta in(0,(pi)/(3))` and `theta in((pi)/(4),(3pi)/(4))`
From case I and II, we have
`theta in((pi)/(4),(pi)/(3))`
Since `((13pi)/(48),(7pi)/(24))in((pi)/(4),(pi)/(3))`
`therefore theta in((13pi)/(48),(7pi)/(24))`
