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5 September, 05:24

2 cos x + 3 sin 2x = 0

answer in degrees

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Answers (1)
  1. 5 September, 07:10
    0
    If want just the approximated solutions in the interval from 0 to 360:

    199.47

    340.53

    90

    270

    If you want all the approximated solutions:

    199.47+360k

    340.53+360k

    90+360k

    270+360k

    Step-by-step explanation:

    2 cos (x) + 3 sin (2x) = 0

    First step: Use double angle identity for sin (2x). That is, use, sin (2x) = 2sin (x) cos (x).

    2 cos (x) + 3*2sin (x) cos (x) = 0

    2 cos (x) + 6sin (x) cos (x) = 0

    Factor the 2cos (x) out, like so:

    2cos (x) [ 1 + 3 sin (x) ]=0

    In order for this product to be zero, we must find when both factors are 0.

    2cos (x) = 0 or 1+3sin (x) = 0

    Let's do 2cos (x) = 0 first.

    2cos (x) = 0

    Divide both sides by 2:

    cos (x) = 0

    So the x-coordinate is 0 on the unit at x=90 deg and x=270 deg (in the first rotation).

    Let's do 1+3sin (x) = 0.

    1+3sin (x) = 0

    Subtract 1 on both sides:

    3sin (x) = -1

    Divide both sides by 3:

    sin (x) = -1/3

    Unfortunately this is not on the unit circle so I'm just going to take sin^-1 or arsin on both sides (this is the same thing sin^-1 or arsin).

    x=arcsin (-1/3) = -19.47 degrees

    So that means - (-19.47) + 180 is also a solution so 19.47+180=199.47.

    And that 360+-19.47 is another so 360+-19.47=340.53.

    So the solutions for [0,360] are

    199.47

    340.53

    90

    270

    If you want all the solutions just add + 360*k to each line where k is an integer.
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