How do I find the solutions of this equation:
$\tan^2 (x)=-1$
How do I find the solutions of this equation?
4 Answers
There is no solution since any real number squared is always positive.
(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 $z$.
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0Thank you for your responses! could you help me in solving this following equation too,$tanx=2x/(1-x^2 )$ – 2011-12-18
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.
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.