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    I have an equation: cosecx - 8cosx = 0

    and I have to solve it, 0 < x < pi

    can I change it to: 1/sinx - 8cosx/sinx = 0

    then get: 1/sinx = 8cosx/sinx

    then: 1 = 8cosxsinx/sinx

    then: 1 = 8cosx


    is this all right????? if not, how do I re-arrange the equation to solve it?
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    (Original post by W.H.T)
    I have an equation: cosecx - 8cosx = 0

    and I have to solve it, 0 < x < pi

    can I change it to: 1/sinx - 8cosx/sinx = 0

    then get: 1/sinx = 8cosx/sinx

    then: 1 = 8cosxsinx/sinx

    then: 1 = 8cosx


    is this all right????? if not, how do I re-arrange the equation to solve it?
    either you posted the question incorrectly or your working is incorrect
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    Wiki Support Team
    (Original post by W.H.T)
    I have an equation: cosecx - 8cosx = 0

    and I have to solve it, 0 < x < pi

    can I change it to: 1/sinx - 8cosx/sinx = 0

    then get: 1/sinx = 8cosx/sinx

    then: 1 = 8cosxsinx/sinx

    then: 1 = 8cosx


    is this all right????? if not, how do I re-arrange the equation to solve it?
    No, you can't put 8cosx over sinx
    I haven't looked into it, but I'd suggest trying to get it in terms of tan, or trying to cancel out either sin or cos....
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    Consider this

     cosec(x) = 8cos(x)

    multiply by sin(x) to get

     8sin(x)cos(x) = 1

    now consider using the identity

     sin(2x) = 2sin(x)cos(x)
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    can I do this instead:

    cosecx - 8cosx = 0

    1/sinx - 8cosx = 0

    1/sinx = 8cosx

    1 = 8cosxsinx

    1 = 4sin2x <------- double angle formulae

    1/4 = sin2x
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    (Original post by spread_logic_not_hate)
    Consider this

     cosec(x) = 8cos(x)

    multiply by sin(x) to get

     8sin(x)cos(x) = 1

    now consider using the identity

     sin(2x) = 2sin(x)cos(x)
    thank you, can you check my post above that its all correct
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    (Original post by Pheylan)
    either you posted the question incorrectly or your working is incorrect
    well thats narrowed it down.
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    Yeah thats looking good to me! I get x = 0.126 + 2n*pi from that
 
 
 
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