Use the identity

`sin(3x)=(-sin^3(x)+3cos^2(x)sin(x))`

`sin(3pi)=(-sin^3(pi)+3cos^2(pi)sin(pi))`

`arcsin(sin(3pi))=arcsin(-sin^3(pi)+3cos^2(pi)sin(pi))`

`=arcsin(0)`

`x=arcsin(0)`

`sin(x)=0`

General solutions for sin(x)=0 are,

`x=0+2pi*n ,pi+2pi*n`

solutions for the range -`pi/2` `<=x<=pi/2` ` `

`x=0`

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Use the identity

`sin(3x)=(-sin^3(x)+3cos^2(x)sin(x))`

`sin(3pi)=(-sin^3(pi)+3cos^2(pi)sin(pi))`

`arcsin(sin(3pi))=arcsin(-sin^3(pi)+3cos^2(pi)sin(pi))`

`=arcsin(0)`

`x=arcsin(0)`

`sin(x)=0`

General solutions for sin(x)=0 are,

`x=0+2pi*n ,pi+2pi*n`

solutions for the range -`pi/2` `<=x<=pi/2` ` `

`x=0`

By definition, `arcsin x = y` if `sin y = x` . Replacing x by sin y, yields:

`arcsin x = arcsin (sin y) = y`

Hence, replacing y by `3pi` , yields:

`arcsin(sin 3pi) = 3pi`

You need to evaluate `arcsin(sin 3pi)` , hence, you need to verify if sin `3pi ` falls in interval [-1,1].

Hence, first you need to evaluate sin 3pi, such that:

`sin 3pi = sin(2pi+pi) = sin 2pi*cos pi + sin pi*cos 2pi = 0*(-1) + 0*1 = 0` Notice that `0 in [-1,1].`

Replacing `sin 3pi` by 0, yields:

`arcsin(sin 3pi) = arcsin 0 = 0, pi, 2pi, 3pi,...,n*pi,..`

**Hence, `arcsin 0 = k*pi,` where `k in Z` , so `arcsin (sin 3pi) = 3pi` .**

Evaluate `arcsin(sin(3pi))`

`arcsin(0)=0`

Be aware that `3pi` is not in the range of the arcsine function.

The range of the arcsine function is `-pi/2<=y<=pi/2.`

By the properties of inverse functions, a function and its inverse function will 'cancel' each other out and result in simply the input of the function. In this problem, we know that `arcsin` and `sin` are inverse functions. Therefore:``

`arcsin(sin(3pi)) = 3pi`

However, if the value is computed in a calculator as written, the result will be zero:

`sin(3pi) = 0 -> arcsin(sin(3pi)) = arcsin(0) = 0`

This is a unique case because of the domain of arcsin(x).

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