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Trignometry

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Joined:	4 Mar 2008
Post:	7

5 Apr 2008 13:17:52 IST

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696

solve

None

cos^-1+cos^-1+cos^-1=3

then pt x⁶+y⁶+z⁶-9/x⁶+y⁶+z⁶=0

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Sairam

Blazing goIITian

Joined: 14 Sep 2007

Posts: 573

5 Apr 2008 16:02:45 IST

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i think u meant cos^-1x+cos^-1y+cos^-1z=3pi
simple
range of cos^-1 is [0,pi]
so each x=y=z=-1
now d remaining is obv.

ashish banga

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Joined: 16 Dec 2006

Posts: 1362

5 Apr 2008 16:07:10 IST

Like 1 people liked this

this is a borderline problem
as cos^-1 x can take maximum value

all terms on L.H.S. have their maximum value
x = y = z = cos

= -1
put to solve cos

is not 1as pointed by elas...

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