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    (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?
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    (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?
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    (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)
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    (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
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    (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
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    (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 :ahee:)

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    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.
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    (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 :ahee:)

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



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    (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...

    Hint : Your answer is 8 less that the actual answer.
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    (Original post by notnek)
    You've missed something from the perimeter...

    Hint : Your answer is 8 less that the actual answer.
    Oh yeah, you're supposed to count the line AB (which is 8) because it asks for the sector. Thanks

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    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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    (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.
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    (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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    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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    (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.
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    (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
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    (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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    (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;


    so if your limits were;

     \displaystyle 0 \leq \theta \leq 2\pi


    if you had to solve  \displaystyle \sin\left(2\theta\right) = 0.4 , notice that you have doubled the angle so the limits must also double, so you have to solve this between;

     \displaystyle 0 \leq 2\theta \leq 4\pi


    if you had to solve  \displaystyle  cos\left(\theta + \frac{\pi}{2}\right) = 0.7 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;

     \displaystyle \frac{\pi}{2} \leq \theta \leq \frac{5\pi}{2}


    and finally, if you had something like  \displaystyle \tan\left(3\theta + \frac{\pi}{6}\right) = 2 you have tripled the angle then added pi/6 to it, so do the exact same to the limits;


     \displaystyle 0 \leq 3\theta \leq 6\pi (after tripling)


     \displaystyle \frac{\pi}{6} \leq 3\theta \leq \frac{37\pi}{6} (these are your new limits!)
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    (Original post by DylanJ42)
    everything that happens to the angle happens to the limits too;


    so if your limits were;

     \displaystyle 0 \leq \theta \leq 2\pi


    if you had to solve  \displaystyle \sin\left(2\theta\right) = 0.4 , notice that you have doubled the angle so the limits must also double, so you have to solve this between;

     \displaystyle 0 \leq 2\theta \leq 4\pi


    if you had to solve  \displaystyle  cos\left(\theta + \frac{\pi}{2}\right) = 0.7 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;

     \displaystyle \frac{\pi}{2} \leq \theta \leq \frac{5\pi}{2}


    and finally, if you had something like  \displaystyle \tan\left(3\theta + \frac{\pi}{6}\right) = 2 you have tripled the angle then added pi/6 to it, so do the exact same to the limits;


     \displaystyle 0 \leq 3\theta \leq 6\pi (after tripling)


     \displaystyle \frac{\pi}{6} \leq 3\theta \leq \frac{37\pi}{6} (these are your new limits!)
    Thank you!!

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