HNC Engineering question help

Hi there,
Mature student here taking on a HNC, being out of education for a number of years I'm finding it hard to get back into the swing of things. I have been posed this question.
One of your commonly-used laboratory instantaneous test signal voltages (vs) is described by the equation…
vs=6sin(2πft - π/4)
Where f = 1MHz 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.
Note: A colleague has reminded you that you need to have your calculator in radians mode (RAD) for this calculation, because the angle is given in radians (i.e. π is featured).
Use suitable software to draw at least two cycles of this signal and annotate the drawing so that non-technical colleagues may understand it.
I believe I have calculated the answer (below) but I am now trying to complete the graph using Desmos. again, I think I have the correct graph but given I dont fully understand what I am looking at I'm finding it difficult to be able to annotate for other to understand. Here is a link to the graph I've drawn up, any help on how I should be annotating it would be greatly appreciated, https://www.desmos.com/calculator/ofovhfemtf
vs=6 sin⁡(2πft-π/4)
First, we divide both sides by 6.
vs/6=sin⁡(2πft-π/4)
Then we take sin¯¹ from both sides. (changes to asin for ease)
asin⁡(vs/6)=2πft-π/4
Then add π/4 to both sides.
asin(vs/6)+π/4=2πft
Finally, we divide both side by 2πf and we have made t the subject.
(asin⁡(vs/6)+π/4))/2πf=t
We now need to input the values to calculate t.
v_s=3V
f=1MHz=1000000Hz=1e^6
(asin(3/6)+π/4)/(2π1e^6 )=t
t=(asin(3/6)+π/4)/2π1e6=1.309/6283185.307=0.0000002083=2.083e^(-7)
t=2.083e^(-7)

That's great thank you! I think I have it, but only if i calculate f as 1 and not 1e6.
v=6sin(2π*1*(2.083e^[-7])-(π/4))
v=-4.243
Then comparing it against the new graph and adding the t value one the slider, this matches the point where it crosses the y axis. Am I close or have I missed the mark again?
https://www.desmos.com/calculator/zddzn4j28f

Hi there,

Mature student here taking on a HNC, being out of education for a number of years I'm finding it hard to get back into the swing of things. I have been posed this question.

One of your commonly-used laboratory instantaneous test signal voltages (vs) is described by the equation…
vs=6sin(2πft - π/4)
Where f = 1MHz 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.
Note: A colleague has reminded you that you need to have your calculator in radians mode (RAD) for this calculation, because the angle is given in radians (i.e. π is featured).
Use suitable software to draw at least two cycles of this signal and annotate the drawing so that non-technical colleagues may understand it.

I believe I have calculated the answer (below) but I am now trying to complete the graph using Desmos. again, I think I have the correct graph but given I dont fully understand what I am looking at I'm finding it difficult to be able to annotate for other to understand. Here is a link to the graph I've drawn up, any help on how I should be annotating it would be greatly appreciated, https://www.desmos.com/calculator/ofovhfemtf

vs=6 sin⁡(2πft-π/4)

First, we divide both sides by 6.
vs/6=sin⁡(2πft-π/4)

Then we take sin¯¹ from both sides. (changes to asin for ease)
asin⁡(vs/6)=2πft-π/4

Then add π/4 to both sides.
asin(vs/6)+π/4=2πft

Finally, we divide both side by 2πf and we have made t the subject.
(asin⁡(vs/6)+π/4))/2πf=t

We now need to input the values to calculate t.
v_s=3V
f=1MHz=1000000Hz=1e^6
(asin(3/6)+π/4)/(2π1e^6 )=t
t=(asin(3/6)+π/4)/2π1e6=1.309/6283185.307=0.0000002083=2.083e^(-7)
t=2.083e^(-7)

That's great thank you! I think I have it, but only if i calculate f as 1 and not 1e6.

v=6sin(2π*1*(2.083e^[-7])-(π/4))
v=-4.243

Then comparing it against the new graph and adding the t value one the slider, this matches the point where it crosses the y axis. Am I close or have I missed the mark again?

https://www.desmos.com/calculator/zddzn4j28f

Welcome to TSR. 😉
A few comment(s) on this post.
Your answer to “Make time (t) the subject of this formula” is indeed good. Good job.
Next, is this a maths or engineering module?
If this is an engineering module, I would recommend that you pay attention to However, if you are not taught about them, you can safely ignore them.
If you have solved the following equation for t,

where the solution t = 2.083 × 10^-7 s (not exact), and by substituting the solution (t = 2.083 × 10^-7 s) into the left-hand side of the equation, you should obtain approximately 3.
https://www.wolframalpha.com/input?i=6*sin%282*pi*1*10%5E6*2.0833*10%5E-7+-+pi%2F4%29
As for the plotting of the graph, you need to input the exact equation into desmos

and there is a need for you to change the scale in the “X-axis” (in case you don’t know)or you would see this in desmos even if you have input the correct equation.Hope it makes sense to you.

This absolutely makes sense now thank you so much! I've figured it out now and can see that where the wave reaches approximately 3 volts it is 2.083e-7 pm the time axis!

Good that you resolve the issue.
It may be late for the suggestion - it would be good that you start the time from t = 0 as time before 0 has no physically meaning in most of the engineering problems.

how to do this ?

This absolutely makes sense now thank you so much! I've figured it out now and can see that where the wave reaches approximately 3 volts it is 2.083e-7 pm the time axis!

Hi, almost finished the assignment but getting mixed results from Desmos 🙈 did you get there in the end?

I’ve got a similar question and I’ve just started my HNC, I’ve worked out t but I’m struggling to make the graph, can I see what you inputted into each box to compare with mine please