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Mathematics 20 Online
OpenStudy (anonymous):

how can i solve sin(7x)=sin(5x)???? is it correct to do sin(5x+2x)=sin(5x)??

OpenStudy (anonymous):

how do you get rid of a "sin" function i an equation?

OpenStudy (anonymous):

in*

OpenStudy (anonymous):

.... sin^-1....???

OpenStudy (anonymous):

so do that to both sides of the equation and see what you get

OpenStudy (anonymous):

x=0?

OpenStudy (anonymous):

after you get rid of the "sin" in the equation 'sin(7x)=sin(5x)', what are you left with?

OpenStudy (anonymous):

7x=5x??

OpenStudy (anonymous):

whats the only number that makes that true?

OpenStudy (anonymous):

You can't get rid of a sin function in an equation btapia... It's a function.

OpenStudy (anonymous):

oh man... this is hard...........

OpenStudy (anonymous):

0 works... Are you in Precalc?

OpenStudy (anonymous):

yea the only number that makes '7x=5x' true is 0 true? or no?

OpenStudy (anonymous):

but the graph shows a lot of intersection......

OpenStudy (anonymous):

No, Do you have an interval over which to solve this?

OpenStudy (anonymous):

-2pi to 2pi

OpenStudy (anonymous):

Don't listen to btapia he has no idea what he is talking about.

OpenStudy (anonymous):

but i have to do it algebraically.....

OpenStudy (anonymous):

how did u get from sin7x=sin5x to 5x+2pi(k)=7x??? wat did u do??

OpenStudy (anonymous):

You know the period is 2pi right?

OpenStudy (anonymous):

Sorry I didn't mean to delete that last comment...

OpenStudy (anonymous):

so u just get rid of the sin and add 2kpi in it.......??

OpenStudy (anonymous):

You take the sin(5x)=sin(7x) and add the period of sin(5x), which is 2pi(k) which is 5x+2pi(k) and set it equal to the sin(7x) giving you 5x+2pi(k)=7x 2x=2pi(k) x=pi(k) So it intersects every pi(k).

OpenStudy (anonymous):

So at -2pi, pi, 0, pi, 2pi

OpenStudy (phi):

Here's one way. It relies on a trick, and there might be a nicer way, but here goes. sin(7x)-sin(5x)= 0 rewrite 7x as 6x+x, and 5x as 6x-x sin(6x+x)- sin(6x-x)=0 expand the sin(6x+x) = sin(6x)cos(x)+cos(6x)sin(x) expand -sin(6x-x)= -sin(6x)cos(x)+cos(6x)sin(x) add the 2 equations to get 2cos(6x)sin(x)=0 or cos(6x)sin(x)=0 now find all the x values in your interval that make either cos(6x)=0 or sin(x)=0

OpenStudy (anonymous):

oh wow i get it now~~~ thank you~~

OpenStudy (anonymous):

phi's way works too, but seriously don't listen to btapia...

OpenStudy (anonymous):

yeah~~ i can sleep now~~~ XDDD

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