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Calculus grade 12

posted by on .

solve for (<_ = less than or equal to / pie = pie sign / -pie = negative pie)

3 sin²x = cos²x ; 0 <_ x < 2pie
cos²x - sin²x = sinx ; -pie < x <_ pie

  • Calculus grade 12 - ,

    1st one:

    3 sin²x = cos²x ; 0 ≤ x < 2π
    sin²x /cos²x = 1/3
    tan^2 x = 1/3
    tanx = ± 1/√3
    set your calculator to radians ....
    x = .5236
    x = π - .5236 = 2.61799
    x = π + .5236 = 2.61799
    x = 2π- .5236 = 5.7596

    2nd:
    cos^2 x - sin^2 x = sinx
    1 - sin^2 x - sin^2 x - sinx = 0
    2sin^2 x + sinx - 1 = 0
    (2sinx -1)(sinx + 1) = 0
    sinx = 1/2 or sinx = -1

    if sinx = 1/2
    x = π/6, or x = 5π/6
    if sinx = -1
    then x = =π/2 or 3π/2 ,but -π < x ≤ π
    so x = -π/2

    x = -π/2 , π/6 , 5π/5

  • Calculus grade 12 - ,

    Thank you Reiny! The first question (1st one) your calculations for the 2nd x is wrong. x = x = π + .5236 = 3.665192654 and not 2.61799

  • Calculus grade 12 - ,

    yup, clearly a typo, since the first part of it is correct

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