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    I need help in solving a maths question:

    Solve this equation for x in the interval 0 < x < 2π giving your answers in terms of π.
    3 sec² x = 4 tan² x.

    Here are the answers:
    π/3, 2π/3, 4π/3, 5π/3.

    Here is my attempt (in radiance of course):

    3/cos²x = 4 tan²x
    3 = 4 tan²x cos²x
    3/4 = tan²x cos²x
    3/4= sin²x/cos²x . cos²x
    3/4 =sin²x
    Root 3/2 = sin x
    x = sin-¹(root 3/2)
    x= 1/3π
    Using the CAST method, π-1/3π = 2/3 π.
    Therefore, x = 1/3π, 2/3π.

    Here is my second attempt, this time taking into account the equation: sec² x = 1 + tan ²x
    3(1 + tan²x) = 4 tan² x
    3 + 3 tan² x = 4 tan² x
    4 tan² x - 3 tan² x - 3=0
    tan² x - 3 =0
    tan² x = 3
    tan x = root 3
    x = tan-¹ (root 3)
    x = 1/3π.
    Using the CAST method, 1/3π + π = 4/3 π
    Therefore, x = 1/3π, 4/3π

    If someone could identify where I went wrong it would really be appreciated.
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    (Original post by Nathan R)
    I need help in solving a maths question:

    Solve this equation for x in the interval 0 < x < 2π giving your answers in terms of π.
    3 sec² x = 4 tan² x.

    Here are the answers:
    π/3, 2π/3, 4π/3, 5π/3.

    Here is my attempt (in radiance of course):

    3/cos²x = 4 tan²x
    3 = 4 tan²x cos²x
    3/4 = tan²x cos²x
    3/4= sin²x/cos²x . cos²x
    3/4 =sin²x
    Root 3/2 = sin x
    x = sin-¹(root 3/2)
    x= 1/3π
    Using the CAST method, π-1/3π = 2/3 π.
    Therefore, x = 1/3π, 2/3π.

    Here is my second attempt, this time taking into account the equation: sec² x = 1 + tan ²x
    3(1 + tan²x) = 4 tan² x
    3 + 3 tan² x = 4 tan² x
    4 tan² x - 3 tan² x - 3=0
    tan² x - 3 =0
    tan² x = 3
    tan x = root 3
    x = tan-¹ (root 3)
    x = 1/3π.
    Using the CAST method, 1/3π + π = 4/3 π
    Therefore, x = 1/3π, 4/3π

    If someone could identify where I went wrong it would really be appreciated.
    Both methods are correct, however you are missing solutions as a direct result of your square rooting. For example in the second one;
    tan^2(x)=3 \rightarrow tan(x)=\pm \sqrt{3}

    while you are taking into account the positive quantity, the negative quantity would get your the rest of the solutions. Choose one of your methods, go along with it, and consider both signs; this would get you all the solutions you need in the domain 0&lt;x&lt;2\pi
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    (Original post by RDKGames)
    Both methods are correct, however you are missing solutions as a direct result of your square rooting. For example in the second one;
    tan^2(x)=3 \rightarrow tan(x)=\pm \sqrt{3}

    while you are taking into account the positive quantity, the negative quantity would get your the rest of the solutions. Choose one of your methods, go along with it, and consider both signs; this would get you all the solutions you need in the domain 0&lt;x&lt;2\pi
    Thanks, it seems so obvious now. Your help is really appreciated.
 
 
 
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