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Posted by on Sunday, January 29, 2012 at 9:31am.

sin2x=cos3x then sinx=?(if 0<x<pi/2)

  • trig - , Sunday, January 29, 2012 at 10:17am

    sin 2x = cos3x = sin(pi/2 - 3x)
    Therefore, one solution is
    2x = pi/2 - 3x
    5x = pi/2
    x = pi/10 = 18 degrees

    There is a second solution at x = pi/2, but that is just outside the domain of interest.

  • trig - , Sunday, January 29, 2012 at 10:25am

    Do you mean
    sin^2 x = cos^3 x
    or do you mean
    sin(2x) = cos(3x)
    ?????
    If you mean sin(2x) = cos(3x)
    then
    2 sin x cos x = 4 cos^3 x - 3 cos x

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

    4 sin^2 x + 2 sin x -1 = 0
    let z = sin x
    4 z^2 + 2 z - 1 = 0
    z = [ -2 +/- sqrt (4+16) ] / 8
    z = [ -2 +/- 2 sqrt 5 ] /8
    or
    sin x = (-1 +/- sqrt 5)/4
    if x in quadrant 1 then use + root
    sin x = (sqrt 5 - 1)/4

  • trig - , Sunday, January 29, 2012 at 10:27am

    Do it the way Dr WLS did it, much simpler and more likely correct.

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