How do I find the solutions of this equation?

Question

How do I find the solutions of this equation:
$$\tan^2 (x)=-1$$

Answer 1

There is no solution since any real number squared is always positive.

Answer 2

(Edited after Julian's comment)

There might be complex solutions. In order to find these write

$$\tan z={1\over i}{e^{iz}-e^{-iz}\over e^{iz}+e^{-iz}}={1\over i}{u-1\over u+1}$$

with $u:=e^{2iz}$. With these conventions we have to solve the equation

$$-\left({u-1\over u+1}\right)^2=-1\ ,\quad{\rm i.e.}\quad {u-1\over u+1}=\pm 1\ .$$

As $(u-1)/(u+1)$ cannot be $1$ the last equation has the single solution $u=0$, so that we now have to determine the solutions $z\in{\mathbb C}$ of the equation

$$e^{2iz}\ = 0\ ;$$

but there is no such $z$.

Answer 3

Hint: $$ \tan^{2}(x) = -1 \Rightarrow 1 + \tan^{2}(x) = 0 \Rightarrow \sec^{2}(x) = 0.$$

Oh, on the other hand note that $x^{2} = -1$ has no real solutions.

Answer 4

Notice that $\tan^2(0) = 0$ and for any other $x \neq 0$, $\tan^2(x) > 0$.

If you plot $\tan^2(x)$ and $-1$ you will see that those 2 curves never intersect, because $\tan^2(x) \geq 0$, $x\in{\mathbb R}$.

Therefore there are no real solutions.