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    find to 2dp the values of x in the interval 0<x<2pi such that
    8tanx -3cos x =0
    when 8tan x -3cos = 0 os the same as 3sin^2x +8sinx -3 =0
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    Answers are in bold
    so the quadratic 3sin^2x+8sinx-3 can be written as (3sinx-1)(sinx+3)

    Then make each bracket equal 0
    so 3sinx-1 = 0 sinx+3 = 0
    sinx = 1/3 sinx = -3
    As the graph of sinx has a range 1 to -1 included there sinx cannot equal -3

    Therefore sinx=1/3
    x = arcsin(1/3)
    x = 0.340 (3.d.p)

    Other solutions are obtained from the graph or from the cast diagram but I prefer the graph...
    From the attachment, pi minus the 0.340 (or the full number that is on your calculator which would be more precise) is 2.802 to 3dp using the full number on the calculator
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    (Original post by Hajra Momoniat)
    ...
    Please don't post full solutions - it's against the rules of this forum.

    It would be great if you continue to help here but we prefer to guide students towards the answer.
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    (Original post by notnek)
    Please don't post full solutions - it's against the rules of this forum.

    It would be great if you continue to help here but we prefer to guide students towards the answer.
    O sorry I didn't know xD
 
 
 
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