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# Solving trigonometric equations watch

1. How do i solve sin(2x-35) = -1 in the interval -180<x<180

Using primary and secondary solution.

Is this correct?
sin(2x-35)=-1
2x-35=-90+-360n
2x=-55+-360n
x=-27.5+-120n
2. (Original post by Tynos)
How do i solve sin(2x-35) = -1 in the interval -180<x<180

Using primary and secondary solution.

Is this correct?
sin(2x-35)=-1
2x-35=-90+-360n
2x=-55+-360n
x=-27.5+-120n
You've gone slightly wrong at the end:

but since the question looks for solutions where , then the solution set is

since is out of range for any
3. (Original post by atsruser)
You've gone slightly wrong at the end:

but since the question looks for solutions where , then the solution set is

since is out of range for any
Do i need the other one like (180-x)+-360n?
4. (Original post by Tynos)
Do i need the other one like (180-x)+-360n?
By "other one", I guess you mean the secondary value? In this case, no, since the curve intersects at a single point only.

It would probably help if you drew the graphs on Wolfram alpha or something similar. It should be clear what's going on then.
5. (Original post by atsruser)
By "other one", I guess you mean the secondary value? In this case, no, since the curve intersects at a single point only.

It would probably help if you drew the graphs on Wolfram alpha or something similar. It should be clear what's going on then.
If it did intersect at more that one point, could you show me how to use the secondary solution please?
6. (Original post by Tynos)
If it did intersect at more that one point, could you show me how to use the secondary solution please?
Suppose . To solve this:

1. Find the "principal value" of the angle whose sine is 0.5. This is the angle whose sine is 0.5, and which lies between -180 and +180. (See halfway down Trig Graphs for a picture).

2. Find the general solution for the principal value; due to the periodicity of , we have solutions when we add on any multiple of 360, Hence:

where is a positive, 0, or negative integer.

3. But we also have another possible solution for the original equation, that we can't get to by adding a multiple of 360. If you look at a graph of , you will see that this occurs at . This is called the secondary value.

4. Find the general solution for the secondary value; again, due to the periodicity of , we have solutions when we add on any multiple of 360, Hence:

where is a positive, 0, or negative integer.

5. Put both solutions together:

6. If I'd started with , everything would be very similar, but my PV would be -30, and my SV would be -180+30 = -150. (Check this on a graph of )

7. In general:

where PV is the principal value and SV is the secondary value.

So, let's turn to your more complex example, but altered so we have a PV and SV.

Suppose . Let , so that .

Well, that's nice, since we just solved that problem, with general solution:

which we can tidy up as

If we want solutions in a specific range, then you can extract them from the general solution by adding/subtracting the appropriate multiples of 180, until you start getting values outside the required range.
7. (Original post by atsruser)
Suppose . To solve this:

1. Find the "principal value" of the angle whose sine is 0.5. This is the angle whose sine is 0.5, and which lies between -180 and +180. (See halfway down Trig Graphs for a picture).

2. Find the general solution for the principal value; due to the periodicity of , we have solutions when we add on any multiple of 360, Hence:

where is a positive, 0, or negative integer.

3. But we also have another possible solution for the original equation, that we can't get to by adding a multiple of 360. If you look at a graph of , you will see that this occurs at . This is called the secondary value.

4. Find the general solution for the secondary value; again, due to the periodicity of , we have solutions when we add on any multiple of 360, Hence:

where is a positive, 0, or negative integer.

5. Put both solutions together:

6. If I'd started with , everything would be very similar, but my PV would be -30, and my SV would be -180+30 = -150. (Check this on a graph of )

7. In general:

where PV is the principal value and SV is the secondary value.

So, let's turn to your more complex example, but altered so we have a PV and SV.

Suppose . Let , so that .

Well, that's nice, since we just solved that problem, with general solution:

which we can tidy up as

If we want solutions in a specific range, then you can extract them from the general solution by adding/subtracting the appropriate multiples of 180, until you start getting values outside the required range.

Wow, thanks alot for your time. I have given you + REP !!!

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