1. ## Sine

If you have

$Y = sin (x) - 0. 6$

How could i show the sine curve on my calculator for

X = 0 to 360 $\circ$

Thanks

If you have

$Y = sin (x) - 0. 6$

How could i show the sine curve on my calculator for

X = 0 to 360 $\circ$

Thanks
This is just a regular sine function that has been shifted down the y axis by 0.6 units.

So it peaks at 0.4, troughs at -1.6, and it's mid point is -0.6.

3. is that with X set to 0 to 360

is that with X set to 0 to 360
Throughout the entire interval.

5. so how can i find the two solutions to the equation? in the peaks?

6. $
x= - 4.712
$

$x= 1.5707$

?

so how can i find the two solutions to the equation? in the peaks?
What do you mean by SOLVE it?

You have one equation and two unknowns $x \text{ and } y$. It is not solveable. What is it you are trying to find? I thought you only wanted to graph it.

8. well now ive graphed it - how can i find the 2 solutions of that equation that are between 0 - 360 degrees?

9. What are you trying to solve for? Is this the way the question was worded? Can you give me the question exactly as it appears in your book? I don't know how to help you because I can't figure out what it is you solving for

10. 2 solutions between

$sin(x) - 0.6 = 0$

between 0 and 360degree

2 solutions between

$sin(x) - 0.6 = 0$

between 0 and 360degree

That looks much better.

Graphically the solutions should be where the curve crosses the x-axis (That's where your output equals to zero).
Algebraically, you will need to solve for x by using the $sin^{-1}$ key on your calculator somewhere.

12. Ok, now I see what you need to do. The sin(x) curve has an infinite domain , but comes back to the x axis (where y equals zero)at x = 0 and at x= pi. In the case of this problem, the sine curve is shifted down .06. So it won't = 0 at x=0 or pi, it will be 0 somewhere just larger than 0 and pi. ( I'm guessing probably at .06). As far as from 0 to 2pi ( or in your case 360 deg) you will actually have 3 answers. The solutions are the x values on the curve when y = 0
Does that make sense? Good luck!

13. Originally Posted by chabmgph
That looks much better.

Graphically the solutions should be where the curve crosses the x-axis (That's where your output equals to zero).
Algebraically, you will need to solve for x by using the $sin^{-1}$ key on your calculator somewhere.
Could you explain how to do that please chap

14. Originally Posted by chabmgph
That looks much better.

Graphically the solutions should be where the curve crosses the x-axis (That's where your output equals to zero).
Algebraically, you will need to solve for x by using the $sin^{-1}$ key on your calculator somewhere.
Could you explain how to do that please chap
You can do this graphically if you own a graphing calculator. However, depending on what model you have, the procedure would be different. And I believe there is a section in this forum devoted to calculator use.

To do this algebraically,
$\sin (x) - 0.6 =0$
$\sin (x) =0.6$
$x=\sin^{-1}(0.6)$
Then to find a approximation of x, find the $sin^{-1}$ key on your calculator and calculate what $sin^{-1}(0.6)$ is.

15. How can there be two answers then? because of where the curve crosses the x axis?

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