the number of solution of the equation `1+ sinx.sin^2(x/2) = 0` , in `[-pi , pi ]` , is

Options

A. 0
B. 1
C. 3
D. none of these

Answer

A

Explanation

We have,
`1 + "sin" x "sin"^(2) (x)/(2) = 0`
For `0 le x le pi`, we observe that
` "sin" x "sin"^(2) (pi)/(2) ge 0`
`rArr 1 + "sin" x "sin"^(2) (x)/(2) ge 1`
`rArr 1+ "sin" x "sin"^(2) (x)/(2) ne 0`
`rArr 1+ "sin" x "sin"^(2) (x)/(2) = 0` has no solution in `[0, pi]`
For ` -pi le x lt 0`, we have observe that
`-1 lt "sin" x "sin"^(2) (x)/(2) le 0`
`rArr 0 lt 1 + "sin" x "sin"^(2) (x)/(2) le 1`
`rArr 1 +"sin" x "sin"^(2) (x)/(2) ne 0`
`rArr 1 +"sin" x "sin"^(2) (x)/(2) = 0` has no solution in `[-pi, pi]`
Hence, the given equation has no solution in `[-pi, pi]`