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Mathematics 14 Online
OpenStudy (anonymous):

Solve the following equation, giving the exact solutions which lie in [0, 2π). (Enter your answers as a comma-separated list.) sin(2x) = sin(x)

OpenStudy (anonymous):

I thought the answers were\[ x=0,\pi/3,5\pi/3 \]

OpenStudy (anonymous):

I'm getting cos(x) = 1/2.

OpenStudy (anonymous):

your dividing when you should subtract and factor.

OpenStudy (anonymous):

2sinxcosx-sinx=0--> sinx(2cosx-1)=0

OpenStudy (anonymous):

Why would you do that?

OpenStudy (anonymous):

because if you divide then you are losing x=0 which works.

OpenStudy (anonymous):

Ah, true. Then it looks like you did everything fine then.

OpenStudy (anonymous):

but i am getting incorrect.

OpenStudy (anonymous):

Might it be how you're entering your solution? Maybe leave out the "x=" part and just put the list of numbers.

OpenStudy (anonymous):

i did that originally

OpenStudy (anonymous):

becuase there is and "x=" in front of the box.

OpenStudy (anonymous):

Hmm, all those solutions work, they are in the correct interval...

OpenStudy (anonymous):

Oh, we're missing π!

OpenStudy (anonymous):

jinks

OpenStudy (anonymous):

i thought that at the same time

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

sin(2π)=sin(π)=0. Gah, I feel even more embarrassed than I did before.

OpenStudy (anonymous):

It's been too long since I've done the trig.

OpenStudy (anonymous):

thanks a bunch! I appreciate it. We had a brainstorm going and it was good.

OpenStudy (anonymous):

Have a great weekend!

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