IIT JEE , AIEEE Forum - Mathematics , Trignometry

suhailbmss_sullad

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

then pt x⁶+y⁶+z⁶-9/x⁶+y⁶+z⁶=0

elastiboysai

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

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...