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    the limits are

    0^\circ \leq x < 180^\circ

    cos(3x-10)=-0.4

    i did this for the limits, is this right?


    0^\circ \leq 3x-10 < 180^\circ


    -10^\circ \leq 3x < 170^\circ


    -30^\circ \leq x < 510^\circ
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    (Original post by thefatone)
    i did this for the limits, is this right?
    Nopes. If 0 < 3x-10<180 then +10 < 3x < 180 + 10
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    (Original post by Zacken)
    Nopes. If 0 < 3x-10<180 then +10 < 3x < 180 + 10
    but surely if i have an answer of -10 (just an example) then if i add 10 then divide by 3 then the answer would be in the limit of

    0^\circ \leq x < 180^\circ


    wouldn't it?
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    (Original post by thefatone)
    but surely if i have an answer of -10 (just an example) then if i add 10 then divide by 3 then the answer would be in the limit of

    0^\circ \leq x < 180^\circ


    wouldn't it?

    I dunno what you're saying, but I advise that you look up solving linear equations.

    It's like you're telling me that 3x-10 = 0 \iff 3x = -10
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    (Original post by Zacken)
    I dunno what you're saying, but I advise that you look up solving linear equations.

    It's like you're telling me that 3x-10 = 0 \iff 3x = -10
    not quite

    i'm trying to say that if you have 3x-10=0 then on the left side then on the right side you must add 10

    so this
    0^\circ \leq 3x-10 < 180^\circ

    the answer can be 10 less because you add 10 when you get the final answers
    thus
    -10^\circ \leq 3x < 170^\circ
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    (Original post by thefatone)
    not quite

    i'm trying to say that if you have 3x-10=0 then on the left side then on the right side you must add 10

    so this
    0^\circ \leq 3x-10 < 180^\circ

    the answer can be 10 less because you add 10 when you get the final answers
    thus
    -10^\circ \leq 3x < 170^\circ
    What?

    When you add 10 you get:

    \displaystyle 

\begin{align*}0 < 3x - 10 < 180 &\Rightarrow 0 + 10 < 3x -10 + 10 < 180 + 10\\& \Rightarrow 10 < 3x + 0 < 190 \\&\Rightarrow 10 < 3x < 190 \end{align*}
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    (Original post by Zacken)
    What?

    When you add 10 you get:

    \displaystyle 

\begin{align*}0 < 3x - 10 < 180 &\Rightarrow 0 + 10 < 3x -10 + 10 < 180 + 10\\& \Rightarrow 10 < 3x + 0 < 190 \\&\Rightarrow 10 < 3x < 190 \end{align*}
    i see, so then i multiply by 3 after that?
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    (Original post by thefatone)
    i see, so then i multiply by 3 after that?
    DIVIDE by 3.
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    (Original post by Zacken)
    DIVIDE by 3.
    this is very confusing... but then when i get my values for
    cos(3x-10)=-0.4

    i wouldn't include like values any bigger than 63.3°
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    (Original post by thefatone)
    this is very confusing... but then when i get my values for
    cos(3x-10)=-0.4

    i wouldn't include like values any bigger than 63.3°
    Yes, precisely. (also you wouldn't include values smaller than 3.33)
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    (Original post by Zacken)
    Yes, precisely. (also you wouldn't include values smaller than 3.33)
    wat... that's not working with my logic and "correct" answers

    here have the question and mark scheme

    https://a44694f152ee562e38a9454969d2...%20Edexcel.pdf
    question 4

    mark scheme is in the pages far down
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    (Original post by thefatone)
    wat... that's not working with my logic and "correct" answers

    here have the question and mark scheme

    https://a44694f152ee562e38a9454969d2...%20Edexcel.pdf
    question 4

    mark scheme is in the pages far down
    Urgh, I read your post wrong. I assumed your inequality was right.

    We have: 0 < x < 180 \Rightarrow 3 \times 0 < 3x < 3 \times 180 \Rightarrow 0 < 3x < 540 \Rightarrow -10 < 3x-10 < 530
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    (Original post by Zacken)
    Urgh, I read your post wrong. I assumed your inequality was right.

    We have: 0 < x < 180 \Rightarrow 3 \times 0 < 3x < 3 \times 180 \Rightarrow 0 < 3x < 540 \Rightarrow -10 < 3x-10 < 530
    so you multiply first then takeaway the 10???
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    (Original post by thefatone)
    so you multiply first then takeaway the 10???
    Yeah.
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    (Original post by Zacken)
    Yeah.
    Thanks for clearing that up
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    (Original post by thefatone)
    Thanks for clearing that up
    No problem.
 
 
 
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