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anonymous
 4 years ago
Can anyone help me with this? sin(2x)sin(3x)=0, to solve by x.
anonymous
 4 years ago
Can anyone help me with this? sin(2x)sin(3x)=0, to solve by x.

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precal
 4 years ago
Best ResponseYou've already chosen the best response.0use a graphing calculator to find it roots (aka x  intercepts), did they give you a restriction? if not then you have infinite solutions.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you might try this: \[2\sin(x)\cos(x)=3\cos^2(x)\sin(x)\sin^3(x)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[3\cos^2(x)sin(x)\sin^3(x)2\sin(x)\cos(x)=0\] \[\sin(x)\left ( 3\cos^2(x)\sin^2(x)2\cos(x)\right )=0\] \[\sin(x)\left ( 3\cos^2(x)(1\cos^2(x))2\cos(x)\right) = 0\] \[\sin(x) \left ( 4\cos^2(x)2\cos(x)1\right ) = 0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if course x = 0 work, but we knew that from the start. now solve the quadratic equation in cosine, and the solution to \[4z^22z1=0\] 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
 4 years ago
Best ResponseYou've already chosen the best response.0taking 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
 4 years ago
Best ResponseYou've already chosen the best response.0these 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
 4 years ago
Best ResponseYou've already chosen the best response.0guess we lost viodhora anyway. darn

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0phi provides a much snappier way. use \[\sin(a)\sin(b)=2\cos(\frac{a+b}{2})\sin(\frac{ab}{2})\] and it pops right yout.
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