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Trig equation... 
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#1
Oct1205, 03:27 PM

P: 154

Hey everyone,
I have to solve this equation below: 1+cos(x)+cos(2x)=sin(x)+sin(2x)+sin(3x) After too many simplifications and factorizations, I got to: (I hope it's right tho) (2sinxcosx)(2cosx+1)=cos(x)(1+2cosx) So yeah, I factorized everything pretty much, but what step to take after that, so I can solve this equation ?? Thanks, 


#2
Oct1205, 05:29 PM

HW Helper
P: 1,021

I haven't checked whether your factorization is correct, but assuming it is, you can continue like this: you can cancel out the factors (1+2cosx) and cosx in each side. In order to be allowed to do this, they can't be zero. Check when they are zero and then check whether those values were solutions of the initial problem. After that, all that's left of your equation is 2sinx = 1 which seems easy.



#3
Oct1205, 05:51 PM

P: 154

oh ok, makes sense (sorry i didnt see the canceling out thingy)
thank you! 


#4
Oct1205, 06:04 PM

HW Helper
P: 1,021

Trig equation...
No problem, I hope it works out. It seems to me that you'll get quite a number of solutions



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