Question

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

Answer 1

@ganeshie8

Answer 2

wow, hold on, a certain part of the question isn't coming out right...

Answer 3

"Find all values of x such that sin2x=sinx and \(0\le x \le 2\pi\). (List the answers in increasing order)

Answer 4

there are at least two ways to solve this
il show you the first/fast way quick

Answer 5

\(\sin(2x) = \sin (x)\)

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

Answer 6

if you don't like that method, try using below identity :
\[\sin(2x) = 2\sin x\cos x\]

Answer 7

oh right! that's one of the double angle identities, if I'm not mistaken

Answer 8

Yes
put everything on one side and try factoring

Answer 9

\(\sin 2x = \sin x\)

\(2\sin x\cos x = \sin x\)

\(\sin x(2\cos x-1) = 0\)

use ur favorite zero product property

Answer 10

put everything on one side? So do you mean set this equation to zero?

Answer 11

yes, i did t that already for you

Answer 12

yep, that's what you meant. and oooh yeah you mentioned this before. My favorite indeed/

Answer 13

hmmm.... so upon solving, I see that we can have \(\Large 0, \frac{\pi}{3}, \pi , \frac{5\pi}{3}\)

Answer 14

Am i missing any?

Answer 15

you're missing just one

Answer 16

oh, would that be \(2\pi\)?

Answer 17

Yes, \(x=2\pi\) satisfies \(\sin(x)=0\)

Answer 18

I forgot that in this scenario, it wasn't excluded. Well alright, I think I got it.

Answer 19

good job!

Answer 20

Thank you! You're terrific at explaining things \(\ddot\smile\)

Answer 21

np
im as terrific as you're smart at picking up on these :)