If `x ne (n pi)/(2), n in Z " and "("cos"x)^("sin"^2 x-3 "sin" x + 2) = 1 " Then," x =`

Options

A. `2n pi + (pi)/(2), n in Z`
B. `(2n + 1) pi - (pi)/(2) , n in Z`
C. `n pi + (-1)^(n) (pi)/(2), n inZ`
D. none of these

Answer

D

Explanation

We have,
`x ne (n pi)/(2)`
`rArr "cos" x ne 0, +1`
`therefore ("cos"x)^("sin"^(2)x-3"sin"x+2) = 1`
`rArr "sin"^(2) x- 3"sin" x + 2 = 0`
`rArr "sin" x = 1 " "[because "sin" x ne 2]`
`rArr "cos" x =0`
But, `"cos" x ne 0`
Hence, the given equation has no solution.