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).