You need to find the general solution to equation `tan theta = -1` such that:

`tan theta = -1 => theta = arctan(-1) + npi`

You need to remember the following property such that:

`arctan(-alpha) = -arctan alpha`

Reasoning by analogy yields:

`theta = -arctan(1) + npi => theta = -pi/4 + npi`

You need to find all solutions of equation `tan theta = -1` , over the interval `[0,2pi]` hence, you need to remember that the tangent function has negative values in quadrant 2 and 4, thus the equation has two solutions over `[0,2pi]` such that:

`tan theta = -1 => theta = pi - pi/4 => theta = (3pi)/4` (quadrant 2)

`tan theta = -1 => theta = 2pi - pi/4 => theta = (7pi)/4` (quadrant 4)

**Hence, evaluating the solutions to the given equation over `[0,2pi]` yields `theta = (3pi)/4` and `theta = (7pi)/4` and evaluating the general solution to the given equation yields `theta = -pi/4 + npi.` **