satellite73
  • satellite73
i have a question. why is the solution to \[sin(3x)=\sin(2x)\] \[x=\frac{\pi}{5}, x=\frac{3\pi}{5}\]?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
i mean i worked it out and found this answer, but the last step of my solution required finding \[\cos^{-1}\left (\frac{1-\sqrt{5}}{4}\right)\] so i am wondering if there is a snappy way to do it, because the answer is so neat and clean.
anonymous
  • anonymous
phi
  • phi
@sat my instinct was to take your approach. But looking further, I see sin(a) - sin(b)= 2 cos(0.5(a+b)) sin(0.5(a-b)) gets us to the answer in a snappy way

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anonymous
  • anonymous
@phi, wow it sure does, doesn't it!
anonymous
  • anonymous
for \[\cos x = \frac{1+ \sqrt5}{4}\] and \[\cos x = \frac{1- \sqrt5}{4}\] Since x=pi/5 is a solution to \[\cos x = \frac{1+ \sqrt5}{4}\] x=pi/5+2kpi are also solutions (k is an integer) also, x=-pi/5+2kpi are also solutions. since cos(-x)=cos(x) Similarly since x=3pi/5 is a solution to \[\cos x = \frac{1- \sqrt5}{4}\] x=3pi/5+2kpi are also solutions (k is an integer) also, x=-3pi/5+2kpi are also solutions. since cos(-x)=cos(x) And for sin (x) = 0 we have x = 0 + kpi are solutions. Therefore, the solutions to the equation sin(2x)=sin(3x) are \[x = k \pi\] \[x=2k \pi \pm \frac{\pi}{5}\] and \[x=2k \pi \pm \frac{3\pi}{5}\]
anonymous
  • anonymous
I'm only continuing on satellite73's response in the attached link.
phi
  • phi
@tomas: replace sin^2(x) with 1-cos^2 x you get sat's equation
anonymous
  • anonymous
|dw:1327764356757:dw|
anonymous
  • anonymous
it means 2n\[2n \Pi + -1^{n} \times b\]
anonymous
  • anonymous
phi's answer is short and sweet and makes it clear, but i am still wondering why, when you take \[\cos^{-1}\left(\frac{1-\sqrt{5}}{4}\right)=\frac{\pi}{5}\]
anonymous
  • anonymous
i mean other than the fact that it is true. why is \[\cos(\frac{\pi}{5})\] the same as the solution to \[4x^2-2x-1=0\]

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