If equation `sin^(-1) (4 sin^(20 theta + sin theta) + cos^(-1) (6 sin theta - 1) = (pi)/(2)` has 10 solution for `theta in [0, n pi]`, then find the minimum value of n

Answer

9

Explanation

We must have
`4 sin^(2) theta + sin theta = 6 sin theta -1`
`rArr 4 sin^(2) theta - 5 sin theta + 1 = 0`
`rArr (4 sin theta - 1) (sin theta - 1) = 0`
`rArr sin theta = (1)/(4) " or " sin theta = 1`
`sin theta = 1` is not possible as otherwise
`6 sin theta - 1 = 4 sin^(2) theta + sin theta = 5`
`:. sin theta = (1)/(4)`
For 10 solution, `theta in [0, 9pi] " or " [0, 10 pi]`
Thus, the least value of `n " is " 9`