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

If equation `sin^(-1) (4 sin^(20 theta + sin theta) + cos^(-1) (6 sin

1.	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» Correct Answer - 9 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`

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