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Question

If (sin1x)2+(sin1y)2+(sin1z)2=3π24, then find the minimum value of x+y+z

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Solution


(sin1x)2+(sin1y)2+(sin1z)2=3π24 Is possible only when (sin1x)2=(sin1y)2=(sin1z)2=π24
x=y=z=±1min(x+y+z)=111=3

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