How to solve $\frac12 \sec^2 \frac x2 = 1$ under restricted domain?

Question

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?

Answer 1

The problem here is when you conclude $$tan^2(x/2) = 1 \implies \tan(x/2) = 1$$

What you can conclude is that $$\tan^2(x/2) = 1 \implies \tan (x/2) = \pm 1$$

So you need also to consider which value(s) of $x$ (in the domain) satisfy $\tan(x/2) = -1$.

Answer 2

$$\frac12 \sec^2 \left(\frac x2\right) = 1 \Rightarrow \frac12 \frac{1}{\cos^2 \left(\frac x2\right)} = 1 \Rightarrow \frac12 =\cos^2 \left(\frac x2\right) \Rightarrow \cos \left(\frac x2\right)= \pm \frac{\sqrt2}{2}$$