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29 October, 21:46

Solve the following trigonometric equation for 0to2pi

(a) cos (4x) - 9 sin (2x) + 4 = 0

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Answers (1)
  1. 29 October, 23:19
    0
    x = pi/12, 5pi/12, 13pi/12, 17pi/12.

    Step-by-step explanation:

    Note that cos (4x) = 1 - 2sin^2 (2x)

    Substituting we have:

    1 - 2sin^2 (2x) - 9 sin (2x) + 4 = 0

    2 sin^2 (2x) + 9 sin (2x) - 5 = 0

    (2 sin 2x - 1) (sin 2x + 5) = 0

    sin 2x = 1/2 and sin 2x = - 5.

    sin 2x = 1/2, gives 2x = pi/6, 5pi/6, 13pi/6, 17pi/6.

    There are no solutions to sin 2x = - 5 because the range of sin x is - 1 to + 1.

    So x = pi/12, 5pi/12, 13pi/12, 17pi/12.
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