How do I find the range of arcsin when the range IS affected by translation? This problem has 3x=arcsin(1/2f(x)) but I'm more concerned with the general rules.

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Answer 1 · 2018-08-14T06:59:44

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With $y= f ( x )$ ,,

$3x = arcsin ( y/2 )$ , restrained to $in [ - pi/2, pi/2 ]$

$rArr x in 1/3 [ - pi/2, pi/2 ] = [ - pi/6, pi/6 ]$ .

( If c is added to x, then

$3x = arcsin ( y/2 )$ + 3c , restrained to $in [ - pi/2+ 3c, pi/2 + 3c ]$

So, $x in 1/3 [ - pi/2, pi/2 ] = [ - pi/6, pi/6 ]$ .

Inversely,

$y = 2 sin ( 3x )$ , in [ - 1, 1 ]#, with no restraint on x.

If you like to do inversion to this inversion, what you get is the

piecewise wholesome inversion ( here I insist on the use of

my piecewise wholesome inverse operator $(( sin )^ ( - 1 ))$ .

$x = 1/3 ( sin )^ ( - 1 )( y/2 ) = kpi + ( - 1 ) ^k arcsin ( y/2 )$ ,

$x in [ kpi - pi/6, kpi + pi/6 ], k = 0, +-1, +-2, +-3, ...$ .

For the graph of this wholesome inverse, use $y = 2 sin 3x$ , that

includes the bit, for the given equation.

See graphical depiction, for both.
graph{(x - 1/3 arcsin (y/2 ))(x-pi/6)(x+pi/6)=0}

The sine wave is in-between $y = +-2$ , with x unbounded.
graph{(y - 2 sin (3x))(y-2)(y+2)=0}

So, there is an immediate need to be serious about caring my

piecewise wholesome operator that is relevant to all time-oriented

x giving natural progressive oscillations, and likewise, in rotations

and revolutions, in this Universe. I had discussed this matter many

a time, in my answers,