Unraveling a Tricky Trig Equation: Tips and Tricks for Solving

  • Thread starter Thread starter mohlam12
  • Start date Start date
  • Tags Tags
    Trig
AI Thread Summary
The discussion focuses on solving the trigonometric equation 1+cos(x)+cos(2x)=sin(x)+sin(2x)+sin(3x). After simplification, the equation is factored into (2sinxcosx)(2cosx+1)=cos(x)(1+2cosx). Participants suggest canceling common factors, provided they are not zero, and checking these values against the original equation. The remaining equation simplifies to 2sinx = 1, which is straightforward to solve. This approach is expected to yield multiple solutions.
mohlam12
Messages
153
Reaction score
0
trig equation... :(

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,
 
Last edited:
Physics news on Phys.org
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.
 
oh ok, makes sense (sorry i didnt see the canceling out thingy)
thank you!
 
No problem, I hope it works out. It seems to me that you'll get quite a number of solutions :smile:
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top