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