## anonymous one year ago what are all the solutions to cos(x) = cos(y) ?

1. anonymous

implifying cos(x) = -1cos(y) Multiply cos * x cosx = -1cos(y) Multiply cos * y cosx = -1cosy Solving cosx = -1cosy Solving for variable 'c'. Move all terms containing c to the left, all other terms to the right. Add 'cosy' to each side of the equation. cosx + cosy = -1cosy + cosy Combine like terms: -1cosy + cosy = 0 cosx + cosy = 0 Factor out the Greatest Common Factor (GCF), 'cos'. cos(x + y) = 0 Subproblem 1 Set the factor 'cos' equal to zero and attempt to solve: Simplifying cos = 0 Solving cos = 0 Move all terms containing c to the left, all other terms to the right. Simplifying cos = 0 The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined. Subproblem 2 Set the factor '(x + y)' equal to zero and attempt to solve: Simplifying x + y = 0 Solving x + y = 0 Move all terms containing c to the left, all other terms to the right. Add '-1x' to each side of the equation. x + -1x + y = 0 + -1x Combine like terms: x + -1x = 0 0 + y = 0 + -1x y = 0 + -1x Remove the zero: y = -1x Add '-1y' to each side of the equation. y + -1y = -1x + -1y Combine like terms: y + -1y = 0 0 = -1x + -1y Simplifying 0 = -1x + -1y The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined. The solution to this equation could not be determined.

2. anonymous

Simplifying****

3. anonymous

@MasterGeen no copy and paste please

4. anonymous

YOU WANTED AN ANSWER RIGHT LOL

5. anonymous

cos(x)=cos(y)-->x=2kpi+y and x=2kpi-y

6. anonymous

@amirreza1870 how did you get that?

7. anonymous

Cos and Sin have period of 2pi, i.e. cos(x+2pi)=cos(x) for exemple. WIth that in mind, cos(x) = cos(y) then $$x = y+2n\pi, n\in N$$

8. anonymous

@M4thM1nd what about the other solution?

9. anonymous

the period of cos(x) is 2pi for example cos60=cos420 so the first answer was 2kpi+y then you know that cos(-x)=cos(x) so the second answer is 2kpi-y.

10. anonymous

another solution is due the fact that Cos is an even function, i.e. cos(-x) = cos(x). So... $$x = -y+2n\pi, n\in N$$

11. anonymous

Ok, make sense. But isn't it more like "insight" than actual computation?

12. anonymous

m4th mind helping me after?

13. anonymous

For a more step-by-step solution, cos(x) = cos(y) x = acos(cos(y)) $$x=\pm y+2n\pi$$

14. anonymous

sure @MasterGeen

15. anonymous

but inverse cosine can only gives a value between [0,pi]

16. anonymous

I missed the solution -x = y + (2pi)n initially because didn't realize cos(-x) = cos(x).

17. anonymous

yes, the range of acos(x) is [0, pi], but you get to keep in mind that cos is positive in the first and fourth quadrant and negative in second and third. So, if you evaluate acos(cos(3pi/2)) you will get 3pi/4 for example. This means we have a "refletion" around x = pi, for possible solutions

18. anonymous

Ok. Makes sense. Thank you

19. anonymous

Np ;)