sin3x = sinx Solve for x.


by smallbadwolf
Tags: sin3x, sinx, solve
smallbadwolf
smallbadwolf is offline
#1
Jan27-06, 06:21 PM
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?
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rusassault
rusassault is offline
#2
Jan27-06, 06:41 PM
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
StatusX
StatusX is offline
#3
Jan27-06, 06:45 PM
HW Helper
P: 2,566
2sinx (cos^2(x) + cos^2(x) - 2) = 0

should be

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

smallbadwolf
smallbadwolf is offline
#4
Jan27-06, 06:48 PM
P: 14

sin3x = sinx Solve for x.


Oh okay thanks guys
ehild
ehild is online now
#5
Jan28-06, 12:44 AM
HW Helper
Thanks
P: 9,819
Quote 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
sporkstorms
sporkstorms is offline
#6
Jan28-06, 12:49 AM
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.
mosimane
mosimane is offline
#7
Apr24-10, 12:26 PM
P: 2
Quote Quote by smallbadwolf View Post
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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