Question

How do you solve `cos2x=secx`?

Answer 1

`cos(2x) = 2cos^2(x) - 1`

`2cos^2(x)-1 = 1/cos(x)`

Multiply by cos(x) both side :

`2cos^3(x)-cos(x) = 1`

`=>2cos^3(x) - cos(x) - 1 = 0`

Let's `cos(x) = X`

`=> 2X^3-X-1=0`

We see one solution `X=1`

`cos(x) = 1 => cos(x) = cos(0) => x = 0 + 2kpi` with `k in ZZ`