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## Jamierox4ev3r one year ago Find all values of x such that sin 2x = sin x and ________________.(List the answers in increasing order.)

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1. Jamierox4ev3r

@ganeshie8

2. Jamierox4ev3r

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

3. Jamierox4ev3r

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

4. ganeshie8

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

5. ganeshie8

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

6. ganeshie8

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

7. Jamierox4ev3r

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

8. ganeshie8

Yes put everything on one side and try factoring

9. ganeshie8

$$\sin 2x = \sin x$$ $$2\sin x\cos x = \sin x$$ $$\sin x(2\cos x-1) = 0$$ use ur favorite zero product property

10. Jamierox4ev3r

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

11. ganeshie8

yes, i did t that already for you

12. Jamierox4ev3r

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

13. Jamierox4ev3r

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

14. Jamierox4ev3r

Am i missing any?

15. ganeshie8

you're missing just one

16. Jamierox4ev3r

oh, would that be $$2\pi$$?

17. ganeshie8

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

18. Jamierox4ev3r

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

19. ganeshie8

good job!

20. Jamierox4ev3r

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

21. ganeshie8

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

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