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Answer 1 · 2016-10-14T13:15:08

#x=1/2#

Now
#cos^-1x-sin^-1x=pi/6-------(1)#

using a standard inverse identity ( see proof below)

#cos^-1x+sin^-1x=pi/2-------(2)#

adding eqns. (1) & (2) #sin^-1x # is eliminated

#2cos^-1x=pi/6+pi/2=(8pi)/12=(2pi)/3#

dividing by 2

#=> cos^-1x=pi/3#

#=>x=1/2#

footnote.

proof of:

#cos^-1x+sin^-1x=pi/2#

Let
# cos^-1x=y#

#=>cosy=x#

#=> sin(pi/2-y)=x#

#=>pi/2-y=sin^-1x#
#=>pi/2=sin^-1x+y#

#=>pi/2=sin^-1x+cos^-1x #