How do we find $x$ for this equation? $$\tan(x+1)-\tan(x)=0$$
I tried Newton's method, but I think my initial guess is not appropriate. I started with $x=1/2$, but I failed to end up with the approximated solution. What would the best initial guess be for this problem? Thanks!
On
The graph of $y = \tan(x+1)$ is identical to that of $y=\tan x$, except shifted leftward by one unit. Given their asymptotic nature, these graphs "obviously"(?) never meet.
On
Another attempt..
1)$ y= \tan(x) $, $-π/2 \gt x \lt π/2.$
The function is strictly monotonically increasing in this interval, I.e.
for $-π/2 < x < x+1 \lt π/2$.
No solution.
2) Let $x = π/2 -\epsilon$, with
$0\lt \epsilon \lt 1,$ then $x+1 > π/2.$
We have:
$\tan(x)>0$, and $\tan(x+1)<0.$
No solution.
3) Similarly for $x < - π/2 $, and $x+1 > -π/2.$
4) The case $x \in (kπ -π/2, kπ +π/2)$, $k \in \mathbb{Z}$,
can be reduced to one of the cases 1)-3),
since $\tan(x+kπ) = \tan(x)$.
The equation has no solution.
$$\tan(x+1)-\tan{x}=\frac{\sin1}{\cos{(x+1)}\cos{x}}$$ and since $\sin1\neq0$, we see that our equation has no solutions.