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March 26, 2017

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Solve the equation 2sin(2x) = cos x.

and

Solve the equation
cos^2(2x)–sin^2(2x)= (√3 / 2).

  • Trig - ,

    1st one:

    2sin(2x) = cosx
    4sinxcosx - cosx = 0
    cosx(4sinx - 1) = 0
    cosx = 0 or sinx = 1/4

    for cosx = 0
    x = 90° or 270° OR x = π/2 or 3π/2 radians

    for sinx=1/4
    x = appr 14.5° or 165.5° OR appr .253 or 2.89 radians

  • Trig - ,

    2nd:

    cos^2(2x)–sin^2(2x)= (√3 / 2)
    cos (4x) = √3/2 , using cos (2A) = cos^2 A - sin^2 A

    I know cos 30° = cos 330° = √3/2
    so 4x = 30° or 4x = 330°
    x = 7.5° or x = 82.5°

    the period of cos 4x is 360/4 ° = 90°
    so adding or subtracting multiples of 90 to any of the above solutions, will yield a new solution

    e.g. 7.5 + 90 = 97.5°
    97.5+90 = 187.5° etc

    check for x = 187.5°

    LS = cos^2 375° - sin^2 375°
    = .9330... - .0669...
    = .866025...

    RS = √3/2 = .866025...
    = LS

    (set your calculator to Radians using the DRG key, and get the equivalent answers in radians)

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