Sin3x = sinx Solve for x.

  1. okay this problem stumped me for a while but here is my work for it, and I just got stuck at the end so if any help can be provided thanks in advance.

    sin3x = sinx

    sin(2x + x) = sinx

    sin2x cos x + cos2x sinx = sinx

    2sinx cosx cosx + (2cos^2(x) -1) sinx = sinx

    2sinx cos^2(x) + 2cos^2(x) sinx - sinx = sinx

    2sinx cos^2(x) + 2cos^2(x) sinx - 2sinx = 0

    2sinx (cos^2(x) + cos^2(x) - 2) = 0

    2sinx (2cos^2(x) - 2) = 0

    4sinx(cos^2(x) -1) = 0

    4sinx(-sin^2(x)) = 0

    4sinx = 0 , -sin^2x = 0

    x = 0 , x = 0, pi, 2pi

    Why is it when I graph the equation sin3x - sinx = 0 on my calculator, it comes with 7 solutions when i only have 3?
     
  2. jcsd
  3. solution

    sin(2x+x)-sinx=0

    sin2xcosx + cos2xsinx-sinx = 0

    2sinxcos^2 x + (cos^2 x - sin^2 x) sinx - sinx = 0

    2sinxcos^2 x + cos^2xsinx - sin^3 x -sinx=0

    sinx(2cos^2 x + cos^2 x - sin^2 x - 1) = 0

    sinx(3cos^2x - sin^2x -1) =0

    so u kno sinx= 0, thus 0, 180, and 360 degrees are three of the answers
    then for inside parenthesis

    3cos^2x - sin^2x - 1 = 0

    3cos^2x - (1- cos^2x) - 1 = 0

    3cos^2x - 1 + cos^2x - 1 = 0

    4cos^2x - 2 = 0

    cos^2x = 1/2

    so u kno Cos^-1( plus/minus sq root of 2 / 2) = 45, 315, 35, 225 degrees

    :zzz: loong problems, well sorta
     
  4. StatusX

    StatusX 2,567
    Homework Helper

    2sinx (cos^2(x) + cos^2(x) - 2) = 0

    should be

    2sinx (cos^2(x) + cos^2(x) - 1) = 0
     
  5. Oh okay thanks guys
     
  6. ehild

    ehild 12,459
    Homework Helper
    Gold Member
    2014 Award

    There is a much simpler method to solve problems like that.

    If sin(x) = sin(y) then either

    y=x=2k*pi

    or y=(pi-x)+2k*pi,

    where k is integer (zero included).

    y=3x now, so either

    3x=x+2k*pi --> x = k*pi

    or

    3x=(2k+1)*pi -x -->x=(2k+1)*pi/4

    ehild
     
  7. Maybe you already know this, but just in case, sin[m * pi] = 0 for all m. So, x=m*pi for all m satisfies the (trivial) equation:
    sin[m*pi] = sin[3*m*pi] = 0

    --
    edit: where m is an integer.
     
  8. The problem is that the use of your double angle was incorrect but the idea of solving the proble is correct the thing to do here is
    sin3x=sinx
    implies sin(2x + x)=sinx

    implies sin2xcosx +cos2xsinx - sinx=0

    implies 2sinx.cosx.cosx +cos2xsinx - sinx=0

    implies sinx(2cos^2(x)+ cos2x - 1)=0

    implies sinx(2cos^2(x) + 2cos^2(x) - 1 -1)=0

    implies sinx(4cos^2(x) - 2)=0
    now using the zero product law

    implies sinx=0 or cos^2(x)=1/2

    then the equation will solve to be x=0 + n360 or x=+or- 45 + n360 where n lies in Z or integers from there you will sub in integers the will give you solutions that lie in your domain you draw your graph
     
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