# Confusing question: try and prove that $x -\tan(x) = (2k+1)\frac{\pi}{2}$ has no solution in $[\frac{3\pi}{4},\frac{5\pi}{4}]$

I am trying to show that $x - \tan(x) = (2k+1)\frac{\pi}{2}$ has no solution in $[\frac{3\pi}{4},\frac{5\pi}{4}]$. However, I seem to be stuck as I don't know where to begin.

The only sort of idea is that if I were to draw a graph of $\tan x$ and the lines

$x- \frac{(2k+1)\pi}{2}$, I can see that in the interval $[3\pi/4,5\pi/4]$ the lines intersect $\tan(x)$ near the asymptotes. I can also sort of say that as $\tan (x)$ is a strictly increasing function on $(\pi/2,3\pi/2)$, this means the difference between any two roots of the equation $x- \tan(x)$, one root being to the left of the zero of $\tan(x)$ in here, namely $x=\pi$ and the other to the right of the root, is smallest when we consider the lines $x - \pi/2$ and $x-3\pi/2$.

I can sort of think of something as well to do with the fact that the tangent to $\tan(x)$ at $x=\pi$ is parallel to each of these lines, so the solutions to $\tan(x) = x- \pi/2$, $\tan(x) = x-3\pi/2$ must be sufficiently far away from $\pi$ or rather lie outside $[3\pi/4,5\pi/4]$.

Apart from that, I have no idea how to attack this problem. Can anyone help please?

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Let $f(x) = x -tan(x)$. Taking the first derivative we see that this function is constantly decreasing. The derivative will be negative everywhere besides $x = \pi$. Hence the values live in the interval $[f(\frac{5\pi}{4}),f(\frac{3\pi}{4})]$. Computing those values you see that for every choice of $k$, the number $(2k+1) \frac{\pi}{2}$ is outside the interval. Basically you need to see that it is true for $k = 0, 1, -1$.
I don't understand what you mean by the values live in the interval $[f(5π/4),f(3π/4)]$. Can you elaborate further?? – user38268 Apr 17 '11 at 12:45
Sure, if a continuous function, $f$, is decreasing on a closed interval $[a,b]$, then it's minimum is $f(b)$ and the maximum is $f(a)$. Furthermore $f$, being continuous, obtains every value between it's minimum and maximum. Hence $f$ maps the interval $[a,b]$ onto the interval $[f(b),f(a)]$. Better? – shamovic Apr 17 '11 at 12:51
Yeap that's as plain as night and day. So I get this: draw the horizontal lines $x_1=f(3\pi/4)=\frac{3\pi}{4}-1$ and $x_2=f(5\pi/4)=\frac{5\pi}{4}+1$. Then all I have to show is none of the lines of the form $x=\frac{(2k+1)\pi}{2}$ are in the space between my two horizontal lines $x_1$ and $x_2$, which should be pretty easy. – user38268 Apr 17 '11 at 13:21
Basically yes. You should get that $\frac{3\pi}{2} > f(\frac{3\pi}/4)$ and $\frac{\pi}{2} < f(\frac{5\pi}{4})$. The others will follow – shamovic Apr 17 '11 at 13:34