Solving $\tan x = x + 1$ analytically.

Question

In my books a problem is given as follows:

Prove that least positive value of x satisfying $\tan{x} = x+1$ lies in interval $(\frac{\pi}{4},\frac{\pi}{2} )$.

I tried to solve it as folllow

$$\tan{x} = x+1$$ $$\frac{\sin x}{\cos x} = x+1$$ $$\sin{x}= (x+1)\cos{x}$$

I differentiate the expression I

$$\cos{x}= \cos{x} -(x+1)\sin{x}$$

$$-(x+1)\sin x= 0$$

Here by $\sin{x} \neq 0$ because for this condition $x = \pi n$ which lies outside the interval and $x = -1$ is outright wrong

The book solves this kind of question using graph by studying the intersection of $y = \tan x$ and $y = x+1$.

SO WHY I DON'T WANT TO USE GRAPHS?

Because, the problem is we don't have graphs with us in exams and sometimes it's not feasible to draw graph by hands

So how I may go about these kind of trigonometrical equations?

Answer 1

Let $$f(x)=\tan(x)-x-1$$ Then $$f(\frac{\pi}{4})=-\frac{\pi}{4} <0 \\ \lim_{x \to \frac{\pi}{2}} f(x)=+\infty $$ and use the Intermediate Value Theorem...If you didn't cover that (yet), this is probably what the textbook is trying to argue.

To prove that this is the smallest positive root, you can use the fact that $$f'(x)>0 \mbox{ on } (0, \frac{\pi}{2})$$

As for graphs, you should be able to draw the graphs of $\tan(x)$ and $x+1$ by hand in the exam. If the problem expects you to draw the graphs, the problem is chosen so that it is feasible to draw them by hand.

Answer 2

I am not an expert but this might be one of the way to know the root lie in the given interval or not

Answer 3

You need to show that there is no root in $[0,\frac\pi4]$.

With $f(x):=\tan x-x-1$, you have $f(0)=-1$ and $f(\frac \pi4)=-\frac\pi4$. If you can show that the function is monotonic, you are done.

The first derivative is $\sec^2x-1$, which is non-negative, hence the claim.

Answer 4

Prove that least positive value of $x$ satisfying $\tan{x} = x+1$ lies in interval $(\frac{\pi}{4},\frac{\pi}{2} )$. $\tag{*}$

Denote $$ f(x)=\tan x-x-1. $$ Then $(*)$ reduces the following problems.

• (a)Show that $f$ has no zero in $(0,\pi/4]$;
• (b)Show that $f$ has a zero in $(\pi/4,\pi/2)$.

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For(a), can you give the monotonicity of $f$ in $[0,\pi/4]$? For (b), do you know the intermediate value theorem?

Answer 5

To demonstrate that the solution (if exist) shall be within $\pi /4$ and $\pi /2$ is actually quite straightforward $$ 0 \leqslant x < \infty \quad \Rightarrow \quad 1 \leqslant 1 + x < \infty \quad \xrightarrow{{\tan x = 1 + x}}\quad 1 \leqslant \tan x < \infty $$ and since $tan x$ is increasing for $ 0 \leqslant x$ $$ 1 \leqslant \tan x < \infty \quad \xrightarrow{{\tan x\;\text{increasing}}}\quad \pi /4 \leqslant x < \pi /2 $$ thus the solution shall be within the intersection of the two domains $$ x:\;1 + x = \tan x\quad \Rightarrow \quad x \in \left[ {0,\infty } \right)\; \cap \;\left[ {\pi /4,\pi /2} \right) = \left[ {\pi /4,\pi /2} \right) $$ Then, that a solution exists and is unique is given by the fact that $$ \begin{gathered} \left. {\left( {1 + x} \right)} \right|_{x\, = \;\pi /4}^{} > \left. {\tan x} \right|_{x\, = \;\pi /4}^{} \hfill \\ \left. {\left( {1 + x} \right)} \right|_{x\, = \;\pi /2}^{} < \left. {\tan x} \right|_{x\, = \;\pi /2}^{} \hfill \\ \end{gathered} $$ and that both functions are continuous and strictly increasing in that interval.