The number of values of `theta` in the interval `[-pi/2,pi/2]` and `theta!=(npi)/5` is where ` n=0,+-1,+-2` and `tantheta=cot(5theta)` and `sin(2theta)=cos(4theta)` is

Answer

3

Explanation

`tan theta=cot 5 theta`
`rArr cos 6 theta=0`
`rArr 4 cos^(3)2 theta-3 cos 2theta=0`
`rArr cos 2 theta=0` or `pm sqrt(3)/2` (i)
`sin 2 theta= cos 4 theta`
`rArr 2 sin^(2) 2 theta+sin 2 theta-1=0`
`rArr (2 sin 2 theta-1)(sin 2 theta+1)=0`
`rArr sin 2 theta=-1` or `sin 2 theta=1/2` ...(ii)
From (i) and (ii)
`cos 2 theta=0` and `sin 2 theta=-1`
`rArr 2 theta=- pi/2`
`rArr theta=-pi/4`
or `cos 2 theta = pm sqrt(3)/2, sin 2 theta = 1/2`
`rArr 2 theta=pi/6, (5pi)/6`
`rArr theta=pi/12, (5pi)/12`
`:. theta=- pi/4, pi/12, (5pi)/12`