A little help for a new engineering student

I've jjust started a HNC engineering course, left school a long time ago so my maths is quite rusty. Need help with the following, please. The equation is:

Vs = 5sin (2πft - π/3)

Where f=1 and t represents time.

Make time (t) the subject of this formula and hence determine the first point in time when the instantaneous signal voltage has a magnitude of +3V

So for the next part of the equation I have is

Vs/5 = Sin (2πft - π/3)

Followed by

Sin (Vs/5) = Sin (Sin( sin(2πft - π/3))

Then
Sin (Vs/5) = 2 π ft - π/3

Then
Sin (Vs) + π/3 = 2 π ft

And finally

t= Sin (Vs/5) + 3/2πf

Not sure if I am close or way off. Much help appreciated, and again sorry I'm jumping back into maths after a long time away.

I've jjust started a HNC engineering course, left school a long time ago so my maths is quite rusty. Need help with the following, please. The equation is:

Vs = 5sin (2πft - π/3)

Where f=1 and t represents time.

Make time (t) the subject of this formula and hence determine the first point in time when the instantaneous signal voltage has a magnitude of +3V

So for the next part of the equation I have is

Vs/5 = Sin (2πft - π/3)

The next step is to 'undo' the sine function. This is sin^-1 [remember how we found angles in trig?]

The next step is to 'undo' the sine function. This is sin^-1 [remember how we found angles in trig?]

Which engineering course are you doing? Im also an Electrical Engineering student now pursuing a masters in Microelectronics.

General Engineering HNC

If youve still not got it, post your current attempt?

Vs/5 = sin(2 πft - π/3)

asin(Vs/5) = asin(sin(2πft - π/3))

asin(Vs/5) = 2πft-π/3

asin(Vs/5) + π/3 = 2πft

t = (asin(Vs/5) + π/3)/(2πf)

I can't seem to upload a picture of it or copy and paste in the correct format

Ive used asin (or arcsin) in the above instead of sin^(-1) (its equivalent notation) and used brackets (this is important) but it was about right. So are you ok now?

If you think this is right, I'll be ok for this part of the question.
For the next part I have to draw at least two cycles of this signal and annotate the drawing so that a "non-technical colleague may understand the relevant information it contains" . Not sure how this graph should look. Will it be a simple hyperbola graph?

In my workbook there wasn't an equation to go hand in hand with a graph so I have nothing to crosscheck it with

Two cycles of voltage against time would just be a normal sin() graph. Nothing to do with hyperpolas. Depending on which software you use (desmos?) youll have to be careful about the value of the frequency and make sure the phase shift is ok, but it should be just typing the formula into your plotting software and adjusting the view so that ~2 cycles are displayed.

Drawing maybe drawing by hand, but you could use plotting software to verify as above?

Ok, I can give this a try. Think I can work it out. Can you send me a link of a similar looking graph so I can be certain how it should look, please?

Can you be a little more specific please? I left school a long time ago. Is the formula correct up until this point?

Or is what you mean I replace all the Sin with Sin-1? For example my last part I wrote down should read

t= Sin-1 (Vs/5) + 3/2πf