Solve `1+sinxsin^2x/2=0`

Explanation

`1+sinxsin^2(x/2) = 0`

`=>1+sinx((1-cosx)/2) = 0`

`=>2+sinx-sinxcosx = 0`

`=>2+sinx-(sin2x/2) = 0`

`=>2+sinx = (sin2x)/2`

`=>4+2sinx = sin2x->(1)`

Now, we know, `-1 le sinx le 1`.

`=>-2 le 2sinx le 2`

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