solve:

$$\frac12 \sec^2 \left(\frac x2\right) = 1$$ and domain $x: (-\pi,\pi) \cup (\pi,3\pi)$.

* sec^2 (x/2) = 2
* sec^2 (x/2) can be re-written as tan(x/2)^2 + 1, therefore
* tan^2(x/2) + 1 = 2
* tan^2(x/2) = 1
* tan(x/2) = 1
* x/2 = pi/4, 5pi/4, -pi/4 these fit in the domain
* x = pi/2, 5pi/2, -pi/2

the solution says 3pi/2 and as another angle for x.... the reference angle is then 3pi/4 but how is this possible? tan(pi/4) is negative for tan(3pi/4) 

can someone please explain - also, is my method correct?