Correct Answer - `theta=(2n+1)(pi)/(14), n in Z`
`(1)/(cos theta)+(1)/(cos 3theta)=(1)/(cos 5theta)`
`rArr (cos 3theta+cos theta) cos 5theta = cos 3theta cos theta`
`rArr 2cos 5thetas cos 3theta + 2cos 5 theta cos theta = 2 cos 3theta cos theta`
`rArr cos 8 theta + cos 2 thetas + cos 6theta + cos theta + cos 4 theta = cos 4 theta + cos 2 theta`
`rArr 2 cos 7 theta cos theta = 0`
` rArr cos 7 theta = 0` (as `cos theta = 0` is not possible)
`therefore cos 7 theta = 0`
`therefore 7theta = (2n + 1)(pi)/(2), n in Z` or `theta =(2n+1)(pi)/(14), n in Z`
