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# Solving trigonometric equations- URGENT VERY HARD watch

1. (Original post by Ilovemaths...)
360-30 because its another revolution you add 180 degrees to 180 making 360 degrees. See in my working out, I do not get how that doesn't give you 1/2?
If you still don't understand my last post let me know and I'll post a diagram.
2. (Original post by Ilovemaths...)
the ones I've underlined in red, when divided by 3 they aren't right, I don't know why because i followed the rules of the quadrants properly.
Same with the minuses, I haven't posted that part put if you guys can tell me where I have gone wrong, then i will apply that to th minus part
Sine function is only positive in the first two quadrants - you should have no solutions between 180 and 360 ... [or in those quadrants in further rotations].
3. (Original post by Notnek)
Do you know CAST? How have you taught to use the quadrant diagram?

For another revolution you need to add 360. So to get to that line in the first quadrant a second time you have to go all the way around anticlockwise (360) and then reach the line (+30) which gives you 360 + 30 = 390.
My teacher didn't rally teach it wellm he does make a mistakes often.
So does this mean that if you add 360 degrees to 30 you get 390 right, meaning 390 is your new theta so you add 360 again to 390 getting 750- which is a solution. Is this right?
4. (Original post by Ilovemaths...)
Attachment 727234
You have the wrong values, It is not 330. It is 30 + 360 = 390
And this divided by 3 will give 130. Same with the other, you got wrong values. You need to add 360, not 300
5. (Original post by Notnek)
If you still don't understand my last post let me know and I'll post a diagram.
If you could post a diagram for finding the negative values for the -1080 part, if you have time please.
6. (Original post by Muttley79)
Sine function is only positive in the first two quadrants - you should have no solutions between 180 and 360 ... [or in those quadrants in further rotations].
I see my mistake thanks, I'm confused on finding the - values.
7. (Original post by Ilovemaths...)
My teacher didn't rally teach it wellm he does make a mistakes often.
So does this mean that if you add 360 degrees to 30 you get 390 right, meaning 390 is your new theta so you add 360 again to 390 getting 750- which is a solution. Is this right?
Yes that's right, there are two solutions less than 360 in the first two quadrants (30 and 150). To get to the next solutions you have to go all the way around again until you reach these two quadrants so you get 360 + 30 = 390 and 360 + 150 = 510.

Then the next solutions will be 360 +390 = 750 and 360 + 510 = 870.
8. (Original post by Notnek)
Yes that's right, there are two solutions less than 360 in the first two quadrants (30 and 150). To get to the next solutions you have to go all the way around again until you reach these two quadrants so you get 360 + 30 = 390 and 360 + 150 = 510.

Then the next solutions will be 360 +390 = 750 and 360 + 510 = 870.
Thank you so much, for the quadrant diagram (for finding the - values) do you use still use quadrant 1 and 2. Also, for quadrant 1 would it be -360-theta or -360+theta
for the 2nd quadrant would it be -180-theta or -180+theta
Alsom for any question you get for solving trig equations, do you add 360 degrees to both the principal values every time providing the range is bigger than 360 degrees.
9. (Original post by Ilovemaths...)
Thank you so much, for the quadrant diagram (for finding the - values) do you use still use quadrant 1 and 2. Also, for quadrant 1 would it be -360-theta or -360+theta
for the 2nd quadrant would it be -180-theta or -180+theta
Alsom for any question you get for solving trig equations, do you add 360 degrees to both the principal values every time providing the range is bigger than 360 degrees.
I'm going out now so can't answer until later. Hopefully someone else will help you before then.
10. (Original post by Notnek)
I'm going out now so can't answer until later. Hopefully someone else will help you before then.
Okay see you later then and thanks again
11. (Original post by Hammad(214508))
You have the wrong values, It is not 330. It is 30 + 360 = 390
And this divided by 3 will give 130. Same with the other, you got wrong values. You need to add 360, not 300
Okay then what do you do to find the - values?
12. guys how do you set out the quadrant diagram to find the negative values????
13. draw out a sine graph from 1080 to -1080 and this will help you better visualise whether the value is positive or negative
14. (Original post by Ilovemaths...)
guys how do you set out the quadrant diagram to find the negative values????
Do you still need help?
15. (Original post by Notnek)
Do you still need help?
Hi, could I just ask, for the quadrant to find the - values, do you go anticlockwise and do you do 180 + theta and 360- theta etc.., then make them negaive, for example, 360-30=330 therefore -390/3=-110. Right.
Lastly, do you add 360 degrees to the quadrants in relation to what the question asks you to find the other values but only if the range is bigger than 360 degrees? Is this correct
Thanks
16. (Original post by Ilovemaths...)
Hi, could I just ask, for the quadrant to find the - values, do you go anticlockwise and do you do 180 + theta and 360- theta etc.., then make them negaive, for example, 360-30=330 therefore -390/3=-110. Right.
Lastly, do you add 360 degrees to the quadrants in relation to what the question asks you to find the other values but only if the range is bigger than 360 degrees? Is this correct
Thanks
This sounds a bit confused. Try it this way:

For negative angles the two relevant quadrants are the same and the acute angle is the same so you can use your same diagram but go clockwise instead of anticlockwise. So you have this

The red angle will be the first negative solution and the blue angle will be the second negative solution.

For the red angle this is 30 degrees after -180 going clockwise so that's -210. Then what's the blue angle?

Then further angles can be found by subtracting 360.

Lastly, do you add 360 degrees to the quadrants in relation to what the question asks you to find the other values but only if the range is bigger than 360 degrees? Is this correct
Yes how far you go depends on the question. If you have a range of -360 to 360 then there's no need to consider angles past 360. But remember that ranges given in a question can change.

E.g. solve sin(2x) = 0.5 -360<x<360

You actually need to consider -720<2x<720

So you would consider solutions between -720 and 720 then divide them by 2 to find x.
17. (Original post by Notnek)
This sounds a bit confused. Try it this way:

For negative angles the two relevant quadrants are the same and the acute angle is the same so you can use your same diagram but go clockwise instead of anticlockwise. So you have this

The red angle will be the first negative solution and the blue angle will be the second negative solution.

For the red angle this is 30 degrees after -180 going clockwise so that's -210. Then what's the blue angle?

Then further angles can be found by subtracting 360.

Yes how far you go depends on the question. If you have a range of -360 to 360 then there's no need to consider angles past 360. But remember that ranges given in a question can change.

E.g. solve sin(2x) = 0.5 -360<x<360

You actually need to consider -720<2x<720

So you would consider solutions between -720 and 720 then divide them by 2 to find x.
What you did was -180-30=-220 degrees, can you not do 180+30=210 then add a minus sign in front of it making -210 degrees?
Would this work for all the solutions and any question?
18. (Original post by Ilovemaths...)
What you did was -180-30=-220 degrees, can you not do 180+30=210 then add a minus sign in front of it making -210 degrees?
Would this work for all the solutions and any question?
What do you mean by "this"? What procedure are you talking about here. In other words, how would you know that you should add 180 and then make it negative?

Of course if you were in different quadrants then adding 180 and making it negative might not be a solution.

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