Question

Question #39aa9

Answer 1

`x=pi/4+k * pi` for all `k in ZZ`
(see below for possible additional solution)

Explanation:

If we assume that `"cost"(x) != 0`
then
`color(white)("XXX")"cost"(x)="cost"(x)xxtan(x)`
implies
`color(white)("XXX")tan(x)=1`

`color(white)("XXX")rarr x=pi/4`, for `x` in Q I and `pi/4+kpi` i general.

With no indication of what the `cost` function is, it seems reasonable to assume `"cost"(x)!=0` in order to derive at least a partial solution.

However, you may have meant `cos(x)` instead of `"cost"(x)`
in which case we should allow for
`color(white)("XXX")cos(x)=0`
with the general solutions
`color(white)("XXX")x=pi/2+k * pi, k in ZZ`