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1. (Original post by kiiten)
If the angle is 2/3 pi and i use the red arrow the answer is wrong - i get 1/6 pi. But if you said the angle starts from 0 shouldnt it be from the green arrow because the angle stops after the bound?

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Can you please post the question and explain your problem with reference to the question?
2. (Original post by DylanJ42)
yes always measure angles from the green arrow

what notnek is saying is that solutions between from where the red arrow starts until a full rotation later (ie 1.25 cirlces) are the only valid ones
Right I see, thanks for clearing that up. What about if the angle was in the A and C quadrant (angle less than 1/2 pi). But the bound starts from 1/2 pi to 5/2 pi. you wouldn't count the angle in the A quadrant because its before 1/2 pi right? But, the angle in the c quadrant would that be calculated from 0 like the green arrow?
3. (Original post by notnek)
Can you please post the question and explain your problem with reference to the question?
I posted the question earlier in this thread (2 pictures)
4. (Original post by kiiten)
Right I see, thanks for clearing that up. What about if the angle was in the A and C quadrant (angle less than 1/2 pi). But the bound starts from 1/2 pi to 5/2 pi. you wouldn't count the angle in the A quadrant because its before 1/2 pi right? But, the angle in the c quadrant would that be calculated from 0 like the green arrow?
yes that is all correct, but also remember that the angle in the A quadrant would be counted when the arrow comes around again, I will do an example to show this as its hard to explain in words
5. (Original post by DylanJ42)
yes that is all correct, but also remember that the angle in the A quadrant would be counted when the arrow comes around again, I will do an example to show this as its hard to explain in words
I understand - you mean after you pass 2 pi your back to the A quadrant but this time its from 2 pi not 0
6. (Original post by kiiten)
I understand - you mean after you pass 2 pi your back to the A quadrant but this time its from 2 pi not 0
yea yea yea thats perfect but ive already done an example so i may as well post it (just to show off my fancy markers )

7. I come to this thread armed with advice about food, toys, training, vet's bills and so on. Then I notice the double "I" in the title.

I now make a hasty exit.
8. (Original post by DylanJ42)
yea yea yea thats perfect but ive already done an example so i may as well post it (just to show off my fancy markers )

Nice diagram thanks
9. Question 2.b) - ive attached my working the correct answer is 49.9

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Attached Images

10. (Original post by kiiten)
Question 2.b) - ive attached my working the correct answer is 49.9

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You've missed something from the perimeter...

11. (Original post by notnek)
You've missed something from the perimeter...

Oh yeah, you're supposed to count the line AB (which is 8) because it asks for the sector. Thanks

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12. If you solve the equation 3sin^2x +7sinx - 6 =0 in the interval 0 < x < 2pi then you let u =sinx You get u = 2/3, which you change to x by doing inverse sin. Do you change the bounds on the cast circle too?

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13. (Original post by kiiten)
If you solve the equation 3sin^2x +7sinx - 6 =0 in the interval 0 < x < 2pi then you let u =sinx You get u = 2/3, which you change to x by doing inverse sin. Do you change the bounds on the cast circle too?

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No keep the bounds the same.

If u = 2/3 then solve sin(x) = 2/3 for 0 < x < 2pi.
14. (Original post by notnek)
No keep the bounds the same.

If u = 2/3 then solve sin(x) = 2/3 for 0 < x < 2pi.
Right so when would you know to change the bounds or not?

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15. Like this question:

3cos^2x=cosx
So u =0 and x = cos^-1 (0)
= 1/2 pi

The bounds are still 0 and 2 pi - do you change it?
I only got 2 of the answers 1.57 and 4.71

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16. (Original post by kiiten)
Right so when would you know to change the bounds or not?

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That's when you have something like sin(2x) = 0.3, 0 < x < 2pi

The angle of the sine is 2x which is different to x (we know the bounds for x)

The bounds for 2x are 0 < x < 4pi.
17. (Original post by kiiten)
Like this question:

3cos^2x=cosx
So u =0 and x = cos^-1 (0)
= 1/2 pi

The bounds are still 0 and 2 pi - do you change it?
I only got 2 of the answers 1.57 and 4.71

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You're missing a solution for u.

Sub u = cos(x):

3u^2 = u

3u^2 - u = 0

u(3u - 1) = 0

So u = 0 or u = 1/3
18. (Original post by notnek)
That's when you have something like sin(2x) = 0.3, 0 < x < 2pi

The angle of the sine is 2x which is different to x (we know the bounds for x)

The bounds for 2x are 0 < x < 4pi.
So only when the equation affects x and not sin/cos/tan

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19. (Original post by kiiten)
So only when the equation affects x and not sin/cos/tan

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everything that happens to the angle happens to the limits too;

if you had to solve , notice that you have doubled the angle so the limits must also double, so you have to solve this between;

if you had to solve then you have added pi/2 to the angle so do the same for the limits, so you'll want to solve the above for;

and finally, if you had something like you have tripled the angle then added pi/6 to it, so do the exact same to the limits;

(after tripling)

20. (Original post by DylanJ42)
everything that happens to the angle happens to the limits too;

if you had to solve , notice that you have doubled the angle so the limits must also double, so you have to solve this between;

if you had to solve then you have added pi/2 to the angle so do the same for the limits, so you'll want to solve the above for;

and finally, if you had something like you have tripled the angle then added pi/6 to it, so do the exact same to the limits;

(after tripling)

Thank you!!

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