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If sinx+siny+sinz=3, x,y,z belong to [0,2π],then, a.x 2 +y 2 +z 2 -xy-yz-zx=0 b.x³+y³-z³=π³/8 c.x³+y³+z³=0 d.x+y+z=0

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

Grade:12th Pass

1 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

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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