Question

what are all the solutions to cos(x) = cos(y) ?

Answer 1

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.

Answer 2

Simplifying****

Answer 3

@MasterGeen no copy and paste please

Answer 4

YOU WANTED AN ANSWER RIGHT LOL

Answer 5

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

Answer 6

@amirreza1870 how did you get that?

Answer 7

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
\(x = y+2n\pi, n\in N\)

Answer 8

@M4thM1nd what about the other solution?

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

m4th mind helping me after?

Answer 13

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

Answer 14

sure @MasterGeen

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

Ok. Makes sense. Thank you

Answer 19

Np ;)