हल कीजिए - `cos theta cos 2 theta cos 3 theta = 1/4`

Explanation

दिया है - `cos theta cos 2 theta cos 3 theta = 1/4`
`rArr 4 cos theta cos 2 theta cos 3 theta = 1`
`rArr (2 cos 3 theta cos theta) 2 cos 2 theta = 1`
`rArr (cos 4 theta + cos 2 theta) 2 cos 2 theta - 1 = 0`
`rArr 2 cos 4 theta cos 2 theta + 2 cos^(2) 2 theta -1 = 0`
`rArr 2 cos 4 theta cos 2 theta + cos 4 theta = 0`
`rArr cos 4 theta (2 cos 2 theta +1) = 0`
यदि `cos 4 theta = 0 " तब " 4 theta = (2n +1) pi/2`
अर्थात `theta = (2n +1) pi/8 , n in Z`
यदि `2 cos 2 theta +1 = 0 rArr cos 2 theta = - 1/2 "तब " 2 theta = 2 n pi pm (2pi)/3`
अर्थात `theta = n pi pm pi/3, n in Z`