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    (Original post by 13 1 20 8 42)
    The first solution set is redundant because all such solutions are contained in the second. The last is incorrect. pi/6 does not work, for instance. For the other solutions you should get cos(x/2) + cos(5x/2) = 0 so 2 * cosx * cos(3x/2) = 0.
    Ok so 2npi/5 is the valid one for the solution only?
    And for the second solution 2cos(x)cos(3x/2) = 0
    => cos(x)cos(3x/2) = 0 so cosx = 0 and cos3x/2 = 0?
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    (Original post by Fatts13)
    Ok so 2npi/5 is the valid one for the solution only?
    And for the second solution 2cos(x)cos(3x/2) = 0
    => cos(x)cos(3x/2) = 0 so cosx = 0 and cos3x/2 = 0?
    Well 4npi/5 is valid but unnecessary
    Indeed. Well, either can be 0 (just clarifying that it's or, not and)
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    (Original post by 13 1 20 8 42)
    Well 4npi/5 is valid but unnecessary
    Indeed. Well, either can be 0 (just clarifying that it's or, not and)
    OMD thank you so very very much, you've been a massive help Sorry for being a hassle.

    I'll rep you once TSR let me XD
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    (Original post by Fatts13)
    OMD thank you so very very much, you've been a massive help Sorry for being a hassle.

    I'll rep you once TSR let me XD
    No worries. Not really a hassle as I am just sitting around today anyway (uni starts tomorrow).
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    (Original post by 13 1 20 8 42)
    No worries. Not really a hassle as I am just sitting around today anyway (uni starts tomorrow).
    If thats the case, here's another question Name:  Screen Shot 2016-10-02 at 20.04.30.png
Views: 22
Size:  20.7 KB
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    (Original post by Fatts13)
    If thats the case, here's another question Name:  Screen Shot 2016-10-02 at 20.04.30.png
Views: 22
Size:  20.7 KB
    first one I got a weird general solution but threw in n = 17 and it worked so I'll assume I did it right. Express everything in terms of sin and cos, remove the denominators, and it should all become clear..

    Second one I have an expression in tan(theta), it's only slightly simpler. Play around with it, express everything in terms of sintheta and costheta again.
    edit: lol it's way simpler just needed to cancel stuff
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    (Original post by 13 1 20 8 42)
    first one I got a weird general solution but threw in n = 17 and it worked so I'll assume I did it right. Express everything in terms of sin and cos, remove the denominators, and it should all become clear..

    Second one I have an expression in tan(theta), it's only slightly simpler. Play around with it, express everything in terms of sintheta and costheta again.
    edit: lol it's way simpler just needed to cancel stuff
    So tan3theta would become sin3theta/cos3theta and cot(7theta +pi/7) would become cos(7theta+pi/7)/sin(7theta + pi/7)?
    And then from there I did:
    cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7) and then opened up sin as well and I'm at loss XD
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    (Original post by Fatts13)
    So tan3theta would become sin3theta/cos3theta and cot(7theta +pi/7) would become cos(7theta+pi/7)/sin(7theta + pi/7)?
    And then from there I did:
    cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7) and then opened up sin as well and I'm at loss XD
    No need to expand like that. Just cross-multiply, get everything on one side and you should recognise the resulting structure.
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    (Original post by 13 1 20 8 42)
    No need to expand like that. Just cross-multiply, get everything on one side and you should recognise the resulting structure.
    so sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7)?
    then does it become tan(3theta)tan(7theta +pi/7)?
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    (Original post by Fatts13)
    so sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7)?
    then does it become tan(3theta)tan(7theta +pi/7)?
    I dunno what is going on there..

    You have  \displaystyle \frac{ \sin 3 \theta}{ \cos 3 \theta} = \frac{ \cos (7 \theta + \frac{\pi}{7})}{ \sin (7 \theta + \frac{\pi}{7})} just cross multiply to get everything off the denominator
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    (Original post by 13 1 20 8 42)
    I dunno what is going on there..

    You have  \displaystyle \frac{ \sin 3 \theta}{ \cos 3 \theta} = \frac{ \cos (7 \theta + \frac{\pi}{7})}{ \sin (7 \theta + \frac{\pi}{7})} just cross multiply to get everything off the denominator
    OMD sorry I meant equals not divide.
    So sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
    I have no clue what to do next
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    (Original post by Fatts13)
    OMD sorry I meant equals not divide.
    So sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
    I have no clue what to do next
    Cos(A + B) = ?
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    (Original post by 13 1 20 8 42)
    Cos(A + B) = ?
    Oh so now we expand? cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7)
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    (Original post by Fatts13)
    Oh so now we expand? cos(7theta +pi/7) = cos(7theta)cos(pi/7) - sin(7theta)sin(pi/7)
    Noo; don't complicate, simplify. Try getting everything on one side of the equation (i.e. something = 0) and it might click..
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    (Original post by 13 1 20 8 42)
    Noo; don't complicate, simplify. Try getting everything on one side of the equation (i.e. something = 0) and it might click..
    My head is gone, it's not clicking anything, been deprived of maths for too long.
    I got sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7) =1
    I honestly don't know how to use cos(A+B) without expanding.
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    (Original post by Fatts13)
    My head is gone, it's not clicking anything, been deprived of maths for too long.
    I got sin(3theta)sin(7theta+pi/7)/cos(3theta)cos(7theta+pi/7) =1
    I honestly don't know how to use cos(A+B) without expanding.
    You had sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
    so cos(3theta)cos(7theta+pi/7) - sin(3theta)sin(7theta+pi/7) = 0
    Isn't this structure familiar?
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    (Original post by 13 1 20 8 42)
    You had sin(3theta)sin(7theta+pi/7) = cos(3theta)cos(7theta+pi/7)
    so cos(3theta)cos(7theta+pi/7) - sin(3theta)sin(7theta+pi/7) = 0
    Isn't this structure familiar?
    Nope I did not know we could do that. I presumed because everything was multiplication so you could only divide and multiply.
    So from there we then have:
    cos(3theta + 7theta + pi/7) = 0
    cos(10theta + pi/7) = 0
    cos^-1(0) = pi/2
    10theta + pi/7 = pi/2 + 2npi
    so then the general solutions are: +/- pi/28 + pin/5? Correct?
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    (Original post by Fatts13)
    Nope I did not know we could do that. I presumed because everything was multiplication so you could only divide and multiply.
    So from there we then have:
    cos(3theta + 7theta + pi/7) = 0
    cos(10theta + pi/7) = 0
    cos^-1(0) = pi/2
    10theta + pi/7 = pi/2 + 2npi
    so then the general solutions are: +/- pi/28 + pin/5? Correct?
    Correct up to cos(10theta + pi/7) = 0 but your solution is incomplete. Note that cosx = 0 whenever x = n*pi + pi/2 (look at the graph).
 
 
 
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