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Sin3x = sinx Solve for x. 
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#1
Jan2706, 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? 


#2
Jan2706, 06:41 PM

P: 1

sin(2x+x)sinx=0
sin2xcosx + cos2xsinxsinx = 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 


#3
Jan2706, 06:45 PM

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 


#4
Jan2706, 06:48 PM

P: 14

Sin3x = sinx Solve for x.
Oh okay thanks guys



#5
Jan2806, 12:44 AM

HW Helper
Thanks
P: 10,767

If sin(x) = sin(y) then either y=x=2k*pi or y=(pix)+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 


#6
Jan2806, 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. 


#7
Apr2410, 12:26 PM

P: 2

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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