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The number of solutions of the equation ...

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

A

zero

B

one

C

two

D

three

Text Solution

Verified by Experts

The correct Answer is:
A
Since , `1+sinx"sin"^(2)(x)/(2)=0`
`therefore1+sinx((1-cosx)/(2))=0`
`rArr2+sinx-sinxcosx=0`
`rArrsin2x-2sinx=4`
which is not possible for any x in `[-pi,pi]`.
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