anonymous
  • anonymous
Can anyone help me with this? sin(2x)-sin(3x)=0, to solve by x.
Mathematics
chestercat
  • chestercat
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precal
  • precal
use a graphing calculator to find it roots (aka x - intercepts), did they give you a restriction? if not then you have infinite solutions.
anonymous
  • anonymous
you might try this: \[2\sin(x)\cos(x)=3\cos^2(x)\sin(x)-\sin^3(x)\]
anonymous
  • anonymous
\[3\cos^2(x)sin(x)-\sin^3(x)-2\sin(x)\cos(x)=0\] \[\sin(x)\left ( 3\cos^2(x)-\sin^2(x)-2\cos(x)\right )=0\] \[\sin(x)\left ( 3\cos^2(x)-(1-\cos^2(x))-2\cos(x)\right) = 0\] \[\sin(x) \left ( 4\cos^2(x)-2\cos(x)-1\right ) = 0\]

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anonymous
  • anonymous
if course x = 0 work, but we knew that from the start. now solve the quadratic equation in cosine, and the solution to \[4z^2-2z-1=0\] is \[z=\frac{1\pm\sqrt{5}}{4}\] so we have \[\cos(x)=\frac{1-\sqrt{5}}{4}\] or \[\cos(x)=\frac{1+\sqrt{5}}{4}\]
anonymous
  • anonymous
taking the inverse cosine of both sides we get \[x = \arccos(\frac{1-\sqrt{5}}{4})=\frac{3\pi}{5}\] and the other one give \[x=\frac{\pi}{5}\]
anonymous
  • anonymous
these numbers are so nice that it leads me to believe there is some other snappier way to find this answer. after all usually you would get some completely random solution, nothing so nice and clean as \[\frac{\pi}{5}\] so there must be something else going on here, but i can't see it right now
anonymous
  • anonymous
guess we lost viodhora anyway. darn
anonymous
  • anonymous
phi provides a much snappier way. use \[\sin(a)-\sin(b)=2\cos(\frac{a+b}{2})\sin(\frac{a-b}{2})\] and it pops right yout.

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