Question

In the function `f(x) = cos - cos^2x` for `-pi <= x <= pi`, how do you find the `x`-intercepts?

Answer 1

Intercepts are at `x=0` or `pi/2` or `-pi/2`

Explanation:

For function `f(x)=cosx−cos^2x` for `-pi<=x<=pi`, intercept on `x` axis will be given where `f(x)=0`.

Hence, we have to solve for `cosx−cos^2x=0` or

`cosx(1-cosx)=0` i.e. either `cosx=0` or `cosx=1`

i.e. `x=0` or `pi/2` or `-pi/2`