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    (Original post by upthegunners)
    then what are we adding on to get these values?
    I think your problem is this CAST rubbish is messing with your head.

    \displaystyle \cos 2x = \frac{1}{2}

    Make a substitution u=2x

    You are looking for 4 positive solutions (the normal cos graph has up to 2 solutions in an interval of 2pi and this is "squashed" - not the proper language but you can think about the transformation yourself).

    \displaystyle \cos u=\frac{1}{2} so your calculator will give you a solution.

    Use the following two rules to find the other solutions:

    \cos u=\cos(2\pi-u)

    \cos u=\cos(u+2\pi)

    Find the four smallest positive solutions.

    Divide your answers by 2 as \displaystyle x=\frac{u}{2}
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    (Original post by Mr M)
    I think your problem is this CAST rubbish is messing with your head.

    \displaystyle \cos 2x = \frac{1}{2}

    Make a substitution u=2x

    You are looking for 4 positive solutions (the normal cos graph has up to 2 solutions in an interval of 2pi and this is "squashed" - not the proper language but you can think about the transformation yourself).

    \displaystyle \cos u=\frac{1}{2} so your calculator will give you a solution.

    Use the following two rules to find the other solutions:

    \cos u=\cos(2\pi-u)

    \cos u=\cos(u+2\pi)

    Find the four smallest positive solutions.

    Divide your answers by 2 as \displaystyle x=\frac{u}{2}
    thanks Mr M.

    But one last thing, do we not only get 3 solutions if we do
    \cos u=\cos(2\pi-u)

    \cos u=\cos(u+2\pi)
    ?

    I am fine with question in the range of 0 to 360. Its the fact it is 2x and we have to double it. I am not sure how to get the values from 0 to 720..

    I am self teaching so I can't just ask the teacher :lol:
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    (Original post by upthegunners)
    thanks Mr M.

    But one last thing, do we not only get 3 solutions if we do
    \cos u=\cos(2\pi-u)

    \cos u=\cos(u+2\pi)
    ?

    I am fine with question in the range of 0 to 360. Its the fact it is 2x and we have to double it. I am not sure how to get the values from 0 to 720..

    I am self teaching so I can't just ask the teacher :lol:
    Remember you should be working in radians not degrees. It is fine to double the interval though.

    You are allowed to use the rules more than once or in combination ...
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    (Original post by upthegunners)
    thanks Mr M.

    But one last thing, do we not only get 3 solutions if we do
    \cos u=\cos(2\pi-u)

    \cos u=\cos(u+2\pi)
    ?

    I am fine with question in the range of 0 to 360. Its the fact it is 2x and we have to double it. I am not sure how to get the values from 0 to 720..

    I am self teaching so I can't just ask the teacher :lol:
    I am going to use degrees just because it is easier to type

    cos2x = 0.5

    arc cos (0.5) = 60 (principal value)

    So 2x = 60

    But arc cos (0.5) has infinite solutions

    Using your preferred method you will find the following

    2x = 60
    2x = 300

    adding 360 will give

    2x = 420
    2x = 660


    From these you get

    x = 30, 150, 210, 330



    the method for radians is identical and can be seen in the attachment earlier in the thread
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    (Original post by Mr M)
    Remember you should be working in radians not degrees. It is fine to double the interval though.

    You are allowed to use the rules more than once or in combination ...
    I always convert to radians at the very end, find it a lot easier to work with :P

    I know, but I am not exactly sure how to
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    (Original post by upthegunners)
    I always convert to radians at the very end, find it a lot easier to work with :P

    I know, but I am not exactly sure how to
    It is a lot more difficult to do that - you think otherwise as you are not yet comfortable with radians. Work with them and you will agree within a day or two.

    So you got 2\displaystyle x=\frac{\pi}{3}

    \displaystyle 2x=2\pi-\frac{\pi}{3}=\frac{5\pi}{3}

    Now add 2\pi to each answer.
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    (Original post by TenOfThem)
    I am going to use degrees just because it is easier to type

    cos2x = 0.5

    arc cos (0.5) = 60 (principal value)

    So 2x = 60

    But arc cos (0.5) has infinite solutions

    Using your preferred method you will find the following

    2x = 60
    2x = 300

    adding 360 will give

    2x = 420
    2x = 660


    From these you get

    x = 30, 150, 210, 330



    the method for radians is identical and can be seen in the attachment earlier in the thread
    TenOfThem, you hero

    Makes since now

    Thanks for that Also, in exams would I still get my marks if I worked in degrees right the way through the question and then converted to radians at the very end? That's what I usually do..

    Also, if my method is not identical to the once in the mark scheme (often its not) Do I still get all the marks for arriving at the correct answer at the end? I have no idea as I have not got a teacher to ask all these things to..
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    (Original post by upthegunners)
    Also, in exams would I still get my marks if I worked in degrees right the way through the question and then converted to radians at the very end?
    Yes but don't do this because it makes you look unintelligent.

    (Original post by upthegunners)
    Also, if my method is not identical to the once in the mark scheme (often its not) Do I still get all the marks for arriving at the correct answer at the end?
    Yes.
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    (Original post by upthegunners)
    TenOfThem, you hero

    Makes since now

    Thanks for that Also, in exams would I still get my marks if I worked in degrees right the way through the question and then converted to radians at the very end? That's what I usually do..

    Also, if my method is not identical to the once in the mark scheme (often its not) Do I still get all the marks for arriving at the correct answer at the end? I have no idea as I have not got a teacher to ask all these things to..

    It is better to work in radians tbh

    Try to get used to it

    Alternative methods are acceptable but you need to practice the standard methods
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    (Original post by Mr M)
    Yes but don't do this because it makes you look unintelligent.
    *worries that post 24 made her look dim*

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    (Original post by TenOfThem)
    *worries that post 24 made her look dim*

    No just lazy.
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    (Original post by Mr M)
    Yes but don't do this because it makes you look unintelligent.



    Yes.
    haha, love the first comment :lol:

    I have just checked your bio Mr M and seen that you are a Maths teacher. (which is awesome)

    I was wondering if you have any 'tips' for succeeding in AS Maths? As I have already said..I don't have a teacher so I am really just depending on myself and I can only hope the way I am doing things will pay off..

    It would just be great to have some 'general tips' from an actual Maths teacher Thanks
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    (Original post by Mr M)
    No just lazy.
    True
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    (Original post by upthegunners)
    haha, love the first comment :lol:

    I have just checked you bio Mr M and seen that you are a Maths teacher. (which is awesome)

    I was wondering if you have any 'tips' for succeeding in AS Maths? As I have already said..I don't have a teacher so I am really just depending on myself and I can only hope the way I am doing things will pay off..

    It would just be great to have some 'general tips' from an actual Maths teacher Thanks
    TenOfThem is a teacher too so you are spoiled for choice.

    No tips - just keep asking questions if you don't understand.
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    (Original post by Mr M)
    TenOfThem is a teacher too so you are spoiled for choice.
    Tell Tale
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    (Original post by Mr M)
    TenOfThem is a teacher too so you are spoiled for choice.

    No tips - just keep asking questions if you don't understand.
    Oh wow haha. Awesome

    I have a Mechanics question here that I know how to do but I am just not exactly sure why I do what I am doing..

    Here goes..

    The acceleration of a particle is . At t=0, P passes thorough a fixed origin, O, with velocity -18m/s.

    i) Find an expression for the velocity at any time t:

    ii) show that P changes direction only once: Okay I know you solve for t and get

    BUT. I am not sure why this shows that it changes direction only once I think I am missing something very obvious.. any ideas?, thanks Mr M.
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    (Original post by upthegunners)
    Oh wow haha. Awesome

    I have a Mechanics question here that I know how to do but I am just not exactly sure why I do what I am doing..

    Here goes..

    The acceleration of a particle is . At t=0, P passes thorough a fixed origin, O, with velocity -18m/s.

    i) Find an expression for the velocity at any time t:

    ii) show that P changes direction only once: Okay I know you solve for t and get

    BUT. I am not sure why this shows that it changes direction only once I think I am missing something very obvious.. any ideas?, thanks Mr M.
    Edit: Sorry didn't read the question properly. I think you are supposed to ignore negative time.
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    (Original post by Mr M)
    Sketch a velocity time graph?
    Would this make it clear I take it? Thanks my friend
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    (Original post by Mr M)
    Edit: Sorry didn't read the question properly. I think you are supposed to ignore negative time.
    I drew the graph but I not sure what I should be looking for that indicates a change in direction...

    I guess it's because the negative time indicates the opposite direction and we only have 2 solutions, one positive and one negative?
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    (Original post by upthegunners)
    Would this make it clear I take it? Thanks my friend
    Yes if you just graph positive t. Alternatively you could integrate to find an expression for displacement and graph that.
 
 
 
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