How do you find the solutions to #sin^-1x=sin^-1(1/x)#?

2 Answers
Ratnaker Mehta
Aug 22, 2016

The Soln. is #x=+-1#.

Explanation:

#sin^-1x=sin^-1(1/x)#

#rArr sin(sin^-1x)=sin(sin^-1(1/x)#

#rArr x=1/x#

#rArr x^2=1#

#rArr x=+-1#

These roots satisfy the given eqn.

Hence, the Soln. is #x=+-1#.

Nghi N.
Aug 22, 2016

#pi/2, (3pi)/2#

Explanation:

arcsin x = arcsin (1/x)
The 2 solutions are:
sin x = 1, and #sin (1/x) = 1#
sin x = - 1 , and #sin (1/x) = - 1#
Any other values of x will make the equations untrue.
a. sin x = 1 --> arc x = pi/2
b. sin x = - 1 --> arc x = (3pi)/2
Answers for (0, 2pi)
#pi/2, (3pi)/2#
Check
arc x = pi/2 --> arcsin (1) = arcsin (1/1)
arc x = (3pi)/2 --> arcsin (-1) = arcsin (1/-1)

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