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If sinx+siny+sinz=3, x,y,z belong to [0,2π],then,
a.x2+y2+z2-xy-yz-zx=0
b.x³+y³-z³=π³/8
c.x³+y³+z³=0
d.x+y+z=0
Hi Menka,
Note than when sinx+siny+sinz = 3
(As we know that sinA can take maximum value of 1)
So each should be 1. Hence sinx=siny=sinz=1.
So in the given interval sinx=siny=sinz=1 is possible only when x=y=z=Π/2
Hence Option (A).
Best Regards,
Ashwin (IIT Madras).
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