Solving for t in a Trig Sin Function

[Sparky_]

Greetings,

I've solved (almost) a problem - with the answer involving sin().

My first solution involves values for which sin() is equal to 0.5 - i.e.30 degrees or 0.5235.

The internal of the sin function is sin(10t -0.927)
t = 0.1451

To complete this solution I need to show the other solutions for this - that is the other values of t for which sin (10t - 0.927) = 0.5.

I thought it would be every pi/2 but I see that does not work.

It's obvious less than pi and greater than pi/2 - meaning the simple sin wave starts at 0 goes through 0.5 crests at 1 at pi/2 and goes back through 0.5 (at what value?) and to 0 at pi.

Thanks so much
-Sparky_

 

[mathman]

sin(pi/6)=sin(5pi/6)=.5

The general form is sinx=sin(pi-x).

I can't figure out the rest of your post.

 

[Sparky_]

I was solving sin(10t-0.927) = 0.5
t = 0.1451

but it will actually be t = 0.1451 + (something periodic)

how do I find the something periodic? - meaning t = 0.1451 =n*pi/2 or some such?

 

[Tedjn]

Suppose you set 10t-0.927 equal to a new variable y. Then you would have sin(y) = 0.5, and

y=\frac{\pi}{6} + 2n\pi \text{ or }y=\frac{5\pi}{6} + 2n\pi

because sine has a period of 2\pi, and \sin a = \sin(\pi - a). Replace y with 10t-0.927 in both of those cases and solve for t.