If `cos x + sqrt(sinx) = 0,` then `x =`

Options

A. `"sin"^(-1) ((sqrt(5)-1)/(2))`
B. `pi-"sin"^(-1) ((sqrt(5)-1)/(2))`
C. `"sin"^(-1) ((1-sqrt(5))/(2))`
D. `"cos"^(-1)( sqrt((sqrt(5)-1)/(2)))`

Answer

A

Explanation

We have, `"cos"x + sqrt("sin" x) = 0`
Clearly, this is meaningful, if `"sin" x ge 0`.
Now,
`"cos" x +sqrt("sin" x ) =0`
`rArr "cos" x = -sqrt("sin" x)`
`rArr "cos"^(2) x = -"sin" x `
`rArr 1-"sin"^(2) x = -"sin" x `
`"sin"^(2) x -"sin" x -1 = 0`
`rArr "sin" x = (-1 +-sqrt(5))/(2)`
`rArr "sin" x = (sqrt(5)-1)/(2) " " [because "sin x le 0]`
`x = "sin"^(-1) ((sqrt(5)-1)/(2))