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- satellite73

i have a question. why is the solution to
\[sin(3x)=\sin(2x)\]
\[x=\frac{\pi}{5}, x=\frac{3\pi}{5}\]?

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- satellite73

- katieb

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- 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

my answer is here
http://openstudy.com/study#/updates/4f23fcc4e4b0a2a9c26655ef

- 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

@phi, wow it sure does, doesn't it!

- 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

I'm only continuing on satellite73's response in the attached link.

- phi

@tomas: replace sin^2(x) with 1-cos^2 x
you get sat's equation

- anonymous

|dw:1327764356757:dw|

- anonymous

it means 2n\[2n \Pi + -1^{n} \times b\]

- 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

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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