# Sin3x = sinx Solve for x.

Tags: sin3x, sinx, solve
 P: 14 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?
 P: 1 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 loong problems, well sorta
 HW Helper P: 2,567 2sinx (cos^2(x) + cos^2(x) - 2) = 0 should be 2sinx (cos^2(x) + cos^2(x) - 1) = 0
 P: 14 Sin3x = sinx Solve for x. Oh okay thanks guys
HW Helper
Thanks
P: 10,767
 Quote by smallbadwolf sin3x = sinx sin(2x + x) = sinx sin2x cos x + cos2x sinx = sinx.....
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
 P: 47 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.
P: 2
 Quote by smallbadwolf 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?
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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