If `sin^2x-2sinx-1=0` has exactly four different solutions in `x in [0,npi]` , then value/values of `n` is/are `(n in N)` 5 (b) 3 (c) 4 (d) 6

Options

A. 5
B. 3
C. 4
D. 6

Answer

A, C

Explanation

`sin^(2) x-2 sin x-1=0`
`rArr (sin x-1)^(2)=2` or `sin x-1 = pm sqrt(2)`
or `sin x=1-sqrt(2)` as `sin x cancel(gt) 1`
there are two solutions in `[0, 2pi]` and two more in `[2pi 4pi]`. Thus, `n=4, 5`.