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    8(sinX)^-2.cosX - 4(cosX)^-2.sinX = 0

    is my derivative


    How do I solve for X easily?
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    (Original post by chrislpp)
    8(sinX)^-2.cosX - 4(cosX)^-2.sinX = 0

    is my derivative


    How do I solve for X easily?
     \displaystyle \frac{8cosx}{sin^2x} - \frac{4sinx}{cos^2x} = 0 \implies  \frac{2cosx}{sin^2x} = \frac{sinx}{cos^2x}

    Multiply both sides by  sin^2xcos^2x , then divide both sides by  cos^3x

    You will get  tan^3 x = 2 \implies tanx = \sqrt[3]2 Solve it to get the solutions.

    Sorry for not replying to your previous thread, i did saw that you quoted me about this question but i was busy so couldn't help.
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    Add 4(cosX)^-2.sinX to both sides (to get an equality), get rid of the fractions, divide one side by the other.
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    (Original post by raheem94)
    .. Solve it to get the solutions.
    This is a bit close to a full solution if you ask me...
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    (Original post by DFranklin)
    This is a bit close to a full solution if you ask me...
    I understand it is close to a full solution. Actually he asked me this question on another thread, but i was busy so can't reply neither did anyone else helped him with it, so i thought that he has been stuck on it for a long time now and as no one was helping him so i decided to give him enough detail to finish off the question.

    Sorry for it though.
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    (Original post by raheem94)
    I understand it is close to a full solution. Actually he asked me this question on another thread, but i was busy so can't reply neither did anyone else helped him with it, so i thought that he has been stuck on it for a long time now and as no one was helping him so i decided to give him enough detail to finish off the question.

    Sorry for it though.
    No worries.

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