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    Hi, I' doing an additional maths past paper and don't know how to do this five mark question:
    Solve the equation 5sin2x=2cos2x in the interval 0<x<360
    Give your answers correct to one decimal place.

    (the more than and less than signs should be including 0 and 360)
    Thank you!
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    (Original post by xxjessietxx)
    Hi, I' doing an additional maths past paper and don't know how to do this five mark question:
    Solve the equation 5sin2x=2cos2x in the interval 0<x<360
    Give your answers correct to one decimal place.

    (the more than and less than signs should be including 0 and 360)
    Thank you!
    so you know the trigonometric identity : sinx / cosx = tanx ?

    use this:

    divide both sides by cos2x

    so 5sin2x/cos2x = 2cos2x/cos2x
    5tan2x = 2
    tan2x = 0.4
    2x = tan^-1(0.4)
    = 21.8 , 201.8, 381.8, 561.8(, 741.8) (degrees)
    (because it's 2x, limit is now 720 degrees)
    x = 10.9, 100.9, 190.9, 280.9(, 370.9)
    because range is from 0-360 degrees, x = 10.9, 100.9, 190.9, 280.9 (degrees)

    hope that helps
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    Thank you that makes sense now xxx
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    (Original post by xxjessietxx)
    Thank you that makes sense now xxx
    No worries How long have you been learning the content for Additional Maths?
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    I started last year but we only have one lesson a week and there's only four of us doing it
 
 
 
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