Solve the equation `cos3t=cos2t`?

`cos2theta=cos3theta.`

`rArr cos3theta-cos2theta=0.`

`rArr -2sin((3theta+2theta)/2)*sin((3theta-2theta)/2)=0.`

`rArr -2sin((5theta)/2)sin(theta/2)=0.`

`rArr sin((5theta)/2)=0, or, sin(theta/2)=0.`

Knowing that, `sinx=o rArr x=kpi, k in ZZ,` we have,

`5/2theta=kpi, or, theta/2=kpi, k in ZZ.`

`theta=2/5kpi, or, theta=2kpi, k in ZZ.`

`rArr theta in {2/5kpi}uu{2kpi}, k in ZZ.`

But, we observe that, `{2kpi : k in ZZ} sub {2/5kpi : k in ZZ},` hence

`"The Soln. Set is "{2/5kpi : k in ZZ}.`

Enjoy Maths.!

We will use here `cosB-cosA=2sin((A+B)/2)sin((A-B)/2)`

Hence, we ccan write `cos2theta=cos3theta` as

`cos2theta-cos3theta=0`

or `2sin((3theta+2theta)/2)sin((3theta-2theta)/2)=0`

or `2sin((5theta)/2)sin(theta/2)=0`

i.e. either `(5theta)/2=npi` or `theta/2=npi`, where `n` is an integer.

i.e. either `theta=(2npi)/5` or `theta=2npi`, where `n` is an integer.

Observe that the latter set (i.e. where `theta=2npi`) is a subset of former set (i.e. where `theta=(2npi)/5`). This is so when `n` is a multiple of `5`.

Hence solution is `theta=(2npi)/5`, where `n` is an integer.

cos 3t = cos 2t
`3t = +- 2t`

a. 3t = 2t
t = 0 or `t = 2kpi`

b. 3t = - 2t
5t = 0 or `5t = 2kpi`
`t = (2kpi)/5`
Check by calculator, with k = 1
`t = (2pi)/5 = 72^@` --> `cos 2t = cos 144^@ = - 0.809` -->
`cos 3t = cos 216^@ = - 0.809`. OK
`t = 2pi` --> `cos 2t = cos 4pi = 1` and `cos 3t = cos 6pi = 1`. OK