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Solve 1 + sin x "sin"^(2) x/2=0....

Solve `1 + sin x "sin"^(2) x/2=0`.

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`1+sin x "sin"^(2) x/2=0`
or `2+2 sin x "sin"^(2) x/2=0`
or `2+sin x(1- cos x)=0`
or `4+2 sin x - sin 2x=0`
or `sin 2x=2 sin x+4`
The above result is not possible for any value of x as L.H.S. has maximum value 1 and R.H.S. has minimum value 2.
Hence, there is no solution.
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