Inverse function of trigonometric inverse function problem y = 2Sin (3x)

y=2sin(3x) y∈[-2,2]
3x=arcsin(y/2)
x=arcsin(y/2)/3
So the inverse function of y = 2Sin (3x) is y = arcsin (x / 2) / 3x ∈ [- 2,2]
The range of the original function is the domain of the inverse function!

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