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    Siustrulka's Avatar
    Siustrulka Posts: 33, Reputation: 1
    Junior Member
     
    #1

    Jan 10, 2008, 08:28 PM
    Solving Trigonometric Equations
    solve:
    3sin^2x-cos^2x=0

    4sin^2x-3=0
    jiten55's Avatar
    jiten55 Posts: 105, Reputation: 8
    Junior Member
     
    #2

    Jan 10, 2008, 11:13 PM
    3sin^2x-cos^2x

    = 3 (1 - cos^2x) -cos^2x
    = 3 - 4 cos^2x

    Hence 4 cos^2x = 3

    cos^2 x = 3/4

    cos x = (+ or -) sqrt (3)/2

    Remember: cos 30 = sqrt (3)/2

    And cos (180 - 30) = - cos 30

    From above: x = 30, 150,.


    In the second case:

    sin^2x = 3/4

    On same principles, sin^2 60 = 3/4, sin^2 (180 +60) = 3/4

    x = 60, 240,.

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