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3 December, 00:02

How to solve sin (3x) = sin (x) in the interval [0,2pi)

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  1. 3 December, 01:07
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    Replace in the right side of the equation sin2 x by (1 - cos2 x), then solve the quadratic equation for cos x.

    cosx - 2=1 - cos2 x

    cos2 x + cosx - 3=0

    Use the new improved quadratic formula (Socratic Search)

    D = d2 = b2 - 4ac=1+12=13 - - > d=±13

    There are 2 real roots:

    cosx = - b 2a ± d 2a = - 12 ± √13 2 =

    Computing by calculator:

    cos x = - 0.5 + 1.80 = 1.30

    cos x = - 0.5 - 1.80 = - 2.30

    Both solutions are rejected (because - 1 < x < 1). The equation is not true.
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