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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:

${sin}^{- 1}$ $sin^(-1)$ values $\in [- \frac{π}{2}, \frac{π}{2}]$ $in [ - pi/2, pi/2 ]$ . and

$x - 2 π \in [- 1, 1] \Rightarrow x \in [- 1 + 2 π, 1 + 2 π]$ $x - 2pi in [ - 1, 1 ]rArr x in [ -1 + 2pi, 1 + 2pi ]$

and, likewise,

$x - 3 π \in [- 1, 1] \Rightarrow x \in [- 1 + 3 π, 1 + 3 π]$ $x - 3pi in [ - 1, 1 ] rArr x in [ - 1 + 3pi, 1 +3pi ]$

and the intersection is null.

The two inverse sines are incompatible.

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