how can i solve sin(7x)=sin(5x)???? is it correct to do sin(5x+2x)=sin(5x)??
how do you get rid of a "sin" function i an equation?
in*
.... sin^-1....???
so do that to both sides of the equation and see what you get
x=0?
after you get rid of the "sin" in the equation 'sin(7x)=sin(5x)', what are you left with?
7x=5x??
whats the only number that makes that true?
You can't get rid of a sin function in an equation btapia... It's a function.
oh man... this is hard...........
0 works... Are you in Precalc?
yea the only number that makes '7x=5x' true is 0 true? or no?
but the graph shows a lot of intersection......
No, Do you have an interval over which to solve this?
-2pi to 2pi
Don't listen to btapia he has no idea what he is talking about.
but i have to do it algebraically.....
how did u get from sin7x=sin5x to 5x+2pi(k)=7x??? wat did u do??
You know the period is 2pi right?
Sorry I didn't mean to delete that last comment...
so u just get rid of the sin and add 2kpi in it.......??
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).
So at -2pi, pi, 0, pi, 2pi
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
oh wow i get it now~~~ thank you~~
phi's way works too, but seriously don't listen to btapia...
yeah~~ i can sleep now~~~ XDDD
Join our real-time social learning platform and learn together with your friends!