Question

How do you solve `cos2theta=costheta`?

Answer 1

`theta epsilon {0^o, 120^o, 240^o}` (if the range is restricted to `[0,2pi)`)

Explanation:

Limiting the range of `theta epsilon [0,2pi)`

Since `cos(2theta)= 2cos^2(theta)-1`
After re-arranging `cos(2theta)=cos(theta)` into the form:
`color(white)("XXXX")` `cos(2theta)-cos(theta)=0`
we can write
`color(white)("XXXX")` `2cos^2(theta)-cos(theta)-1 = 0`
which factors as
`color(white)("XXXX")` `(2cos(theta)+1)*(cos(theta)-1) = 0`

So the equation holds if
`color(white)("XXXX")` `cos(theta)= (-1/2)`
`color(white)("XXXX")` `color(white)("XXXX")` `rarr theta = 120^o or 240^o`
or if
`color(white)("XXXX")` `cos(theta)=1`
`color(white)("XXXX")` `color(white)("XXXX")` `rarr theta = 0^o`