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# Trig equations - Adding/Subtracting cycles watch

1. I was taught the method of adding/subtracting cycles of 180 or 360 for multiple angles however I keep getting mixed up and now I'm completely confused

If for e.g. it said solve for in between

As it is 2x it would be between so when I get my values for 2x I add 360 right?

What about if it was between and for or even

and even for if it was
Unless there is a way of working out how much to add, because I don't really understand why you add 180/360/any others, thanks so much!
2. (Original post by IgorYakov)
I was taught the method of adding/subtracting cycles of 180 or 360 for multiple angles however I keep getting mixed up and now I'm completely confused

If for e.g. it said solve for in between

As it is 2x it would be between so when I get my values for 2x I add 360 right?

What about if it was between and for or even

and even for if it was
Unless there is a way of working out how much to add, because I don't really understand why you add 180/360/any others, thanks so much!
When you have a solution in the form a<kx<b
a/k < x < b/k
This is another way of solving for x. . .
3. (Original post by IgorYakov)
I was taught the method of adding/subtracting cycles of 180 or 360 for multiple angles however I keep getting mixed up and now I'm completely confused

If for e.g. it said solve for in between

As it is 2x it would be between so when I get my values for 2x I add 360 right?

What about if it was between and for or even

and even for if it was
Unless there is a way of working out how much to add, because I don't really understand why you add 180/360/any others, thanks so much!
I would always suggest that you do a quick sketch graph. That should make things quite clear.

Personally, if I was given I would change the variable so that and then you would solve the equation which is more straightforward. Once you have your values for , it is a simple matter to convert back to .
4. Multiply the values in the range by the coefficient near x. Remember that so in this case coefficient is a half - -180<x<180 will become -90<x/2<90.
5. (Original post by joostan)
When you have a solution in the form a<kx<b
a/k < x < b/k
This is another way of solving for x. . .

(Original post by Brister)
I would always suggest that you do a quick sketch graph. That should make things quite clear.

Personally, if I was given I would change the variable so that and then you would solve the equation which is more straightforward. Once you have your values for , it is a simple matter to convert back to .

(Original post by JerzyDudek)
Multiply the values in the range by the coefficient near x. Remember that so in this case coefficient is a half - -180<x<180 will become -90<x/2<90.

But I have to add multiples of 180 or 360 don't I? I don't understand how much to add e.g.

Solve

(2x) = 30 and 150 and
(2x) = 390 and 510 (added 360 to both values as the new range is

but I don't know how to do this for the examples I gave in the first post, for negatives or 180 etc

thanks!
6. (Original post by IgorYakov)
But I have to add multiples of 180 or 360 don't I? I don't understand how much to add e.g.

Solve

(2x) = 30 and 150 and
(2x) = 390 and 510 (added 360 to both values as the new range is

but I don't know how to do this for the examples I gave in the first post, for negatives or 180 etc

thanks!
Okay, since you are working with sine you will be adding/subtracting 360, because that is the period of sine. However, there are two principal solutions which in your case you correctly identify as and . The symmetry of the graph of sine will give you this easily.

If you have to deal with ranges other than then you just keep adding/subtracting 360 until you can no longer find more solutions in that range.

If you are working with , then you can always write . Otherwise you can modify your graph and range accordingly.
7. (Original post by IgorYakov)
But I have to add multiples of 180 or 360 don't I? I don't understand how much to add e.g.

Solve

(2x) = 30 and 150 and
(2x) = 390 and 510 (added 360 to both values as the new range is

but I don't know how to do this for the examples I gave in the first post, for negatives or 180 etc

thanks!
Not necessarily multiples of 180. Just follow what I said - multiply the range by a number near x. Say you've got a range of 0<x<360 and you have to solve an equation which has 2x in brackets (e.g. sin(2x), cos(2x)), this means you have to double x and thus you double 0 and 360 as well: 0<2x<720. Now to check you can divide the whole inequality by 2 and you'll get you original range.
8. (Original post by Brister)
Okay, since you are working with sine you will be adding/subtracting 360, because that is the period of sine. However, there are two principal solutions which in your case you correctly identify as and . The symmetry of the graph of sine will give you this easily.

If you have to deal with ranges other than then you just keep adding/subtracting 360 until you can no longer find more solutions in that range.

If you are working with , then you can always write . Otherwise you can modify your graph and range accordingly.
Ohh okay, so I only ever add or subtract 360, not 180 or 90 or anything else? Or will I need 180 for tan?
9. (Original post by IgorYakov)
Ohh okay, so I only ever add or subtract 360, not 180 or 90 or anything else? Or will I need 180 for tan?
You will indeed need 180 for tan.
10. (Original post by Brister)
You will indeed need 180 for tan.

Okay! Thanks
just to make sure I'm clear on this, if it was 2x for Sin or Cos I would add/subtract 360 so they're inside the range
but if it was 2x for tan I would +/- 180?
11. (Original post by IgorYakov)
Okay! Thanks
just to make sure I'm clear on this, if it was 2x for Sin or Cos I would add/subtract 360 so they're inside the range
but if it was 2x for tan I would +/- 180?
Yes, that is correct.
12. (Original post by Brister)
Yes, that is correct.

Thank you so much!
13. (Original post by Brister)
Yes, that is correct.
Hi, but if the range given was e.g. would you +/- 360 even for Tan?
14. (Original post by Secret.)
Hi, but if the range given was e.g. would you +/- 360 even for Tan?
No dude, changing the range of won't change the behavior of . The period of the tangent function is still 180. The only difference is that you may have less values for which solve the particular equation.
15. (Original post by Brister)
No dude, changing the range of won't change the behavior of . The period of the tangent function is still 180. The only difference is that you may have less values for which solve the particular equation.
Okk thanks!

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