Find the exact solution of 5 sin^-1 (x - 2pi) = 2 sin^-1 (x - 3pi) ?

1 Answer
A. S. Adikesavan
Aug 13, 2018

I invalidate this equation.

Explanation:

#sin^(-1)# values #in [ - pi/2, pi/2 ]#. and

#x - 2pi in [ - 1, 1 ]rArr x in [ -1 + 2pi, 1 + 2pi ]#

and, likewise,

#x - 3pi in [ - 1, 1 ] rArr x in [ - 1 + 3pi, 1 +3pi ]#

and the intersection is null.

The two inverse sines are incompatible.

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