Find all values of x such that sin 2x = sin x and ________________.(List the answers in increasing order.)

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Find all values of x such that sin 2x = sin x and ________________.(List the answers in increasing order.)

Mathematics
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wow, hold on, a certain part of the question isn't coming out right...
"Find all values of x such that sin2x=sinx and \(0\le x \le 2\pi\). (List the answers in increasing order)

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there are at least two ways to solve this il show you the first/fast way quick
\(\sin(2x) = \sin (x)\) since \(\sin(t)\) has a period of \(2\pi\), we must have \[2x=x+2n\pi \implies x = 2n\pi\tag{1}\] since \(\sin(t)=\sin(\pi-t)\), we must have \[2x=\pi-x+2n\pi\implies x =\dfrac{\pi}{3}+\dfrac{2n\pi}{3} \tag{2}\] plugin \(n=0,1,2\ldots\) and take the solutions that are within the given interval
if you don't like that method, try using below identity : \[\sin(2x) = 2\sin x\cos x\]
oh right! that's one of the double angle identities, if I'm not mistaken
Yes put everything on one side and try factoring
\(\sin 2x = \sin x\) \(2\sin x\cos x = \sin x\) \(\sin x(2\cos x-1) = 0\) use ur favorite zero product property
put everything on one side? So do you mean set this equation to zero?
yes, i did t that already for you
yep, that's what you meant. and oooh yeah you mentioned this before. My favorite indeed/
hmmm.... so upon solving, I see that we can have \(\Large 0, \frac{\pi}{3}, \pi , \frac{5\pi}{3}\)
Am i missing any?
you're missing just one
oh, would that be \(2\pi\)?
Yes, \(x=2\pi\) satisfies \(\sin(x)=0\)
I forgot that in this scenario, it wasn't excluded. Well alright, I think I got it.
good job!
Thank you! You're terrific at explaining things \(\ddot\smile\)
np im as terrific as you're smart at picking up on these :)

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