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C3 Trigonometry

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    (Original post by hassi94)
    That's because it's trying to do tan(90) (undefined) then doing 2/it, instead of just doing 2cos(90) then dividing by sin90.
    I see what you did there:adore:
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    (Original post by hassi94)
    210 is, but it seems you've used an incorrect method (just got one correct answer coincidentally).

    Right so if I ask you to write down the values of 3x/2 for the range 0<x<360 what are your answers?
    I got 315,135,45, 495

    Then divided them by 1.5 to get 210,90,45, 330?

    But tan90 is undefined isn't it?
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    tan(3x/2)=-1
    a=3x/2
    tan(a)=-1 => a=3(pi)/4 + k(pi), k is an integer.
    3x/2=3(pi)/4 + k(pi)
    3x=3(pi)/2 + 2k(pi)
    x=(pi)/2 +2k(pi)/3, k is an integer.
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    (Original post by GreenLantern1)
    I got 315,135,45, 495

    Then divided them by 1.5 to get 210,90,45, 330?

    But tan90 is undefined isn't it?
    You shouldn't have got 45 in either bit, it was -45 then +n180 for the first bit

    And that's right otherwise.

    Nowhere is tan90 done, though, because tan(x) isn't anywhere in the equation.
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    (Original post by hassi94)
    You shouldn't have got 45 in either bit, it was -45 then +n180 for the first bit

    And that's right otherwise.

    Nowhere is tan90 done, though, because tan(x) isn't anywhere in the equation.
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
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    (Original post by GreenLantern1)
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
    Yep.

    Now do you still need help with that other question?
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    (Original post by GreenLantern1)
    So is 90 correct or not?

    Please tell me exactly what the correct answers are as I have been confused.

    Is it 210, 90 and 330?
    Do you know the general solution for tangent?

    If you consider the graph of tanx, and x = 0.5,
    http://www.wolframalpha.com/input/?i=plot+tanx+and+0.5
    they meet once in every period.
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    (Original post by hassi94)
    Yep.

    Now do you still need help with that other question?
    Yes please

    I am just clueless as to how to approach it!
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    (Original post by GreenLantern1)
    Yes please

    I am just clueless as to how to approach it!
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

     \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
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    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

     \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    So the angles in answer to your question are x= 0,180,360,66.4,293.6
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    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

     \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    For my one: 2 cot y = 3 cos y

    I get:

    2 cot y = 3 cos y
    2 cot y - 3 cos y = 0
    2(cos y/sin y)- 3 cos y = 0
    cos y ( 2 cosec y -3)= 0
    y=90,270,41.8,132.8?
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    (Original post by hassi94)
    Right I don't really want to just give you the answer so I'll show you a similar example:

    2 \tan x = 5 \sin x

    2 \tan x - 5 \sin x = 0

    2 \dfrac{ \sin x}{ \cos x} - 5 \sin x = 0

     \sin x \left( \dfrac{2}{ \cos x} - 5 \right) = 0

    So either \sin x = 0 OR \dfrac{2}{ \cos x} - 5 = 0

    Then just rearrange the cos one, inverse sine/cos and find solutions in your range.
    I also did another question: 2sin^2x=3cosx and i got values of 60,300 after making it into a quadratic. Is that correct; I think it is?
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    (Original post by GreenLantern1)
    For my one: 2 cot y = 3 cos y

    I get:

    2 cot y = 3 cos y
    2 cot y - 3 cos y = 0
    2(cos y/sin y)- 3 cos y = 0
    cos y ( 2 cosec y -3)= 0
    y=90,270,41.8,132.8?
    Haven't checked the values (and probably won't tonight - I'm quite ill so am going to bed now) but your working is perfect.

    For the next question 2sin^2x = 3cosx 60 and 300 are correct. Well done you're picking this up quite well now
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    (Original post by hassi94)
    Haven't checked the values (and probably won't tonight - I'm quite ill so am going to bed now) but your working is perfect.

    For the next question 2sin^2x = 3cosx 60 and 300 are correct. Well done you're picking this up quite well now
    Thanks, I was just a bit stuck at first

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