Find the roots of the function

Question

find the roots of the function when $x$ goes from $0$ to $3$?

$$f(x) = \tan(2 x) - 1$$

I tried to set $f(x)=0$ but came up with $x=28$

Answer 1

Just set $f(x) = 0$, that is,

$\tan(2x) - 1 = 0 \implies \tan(2x) = 1 \implies 2x = \arctan(1) \implies x = \frac{\arctan(1)}{2} \approx 0.3927$

Answer 2

$\tan$ is a $\pi$-periodic function and $\tan \frac\pi 4 = 1$. Therefor $$\begin{align*} \tan(2x) & = 0 \\ \Leftrightarrow 2x & = \frac \pi 4 + k\pi & k\in\mathbb Z\\ \Leftrightarrow x & = \frac \pi 8 + k\frac\pi 2 & k\in\mathbb Z \end{align*}$$ For $x\in[0,3]$ you get $k\in\{0, 1\}$ and thus $$x \in \{\frac\pi 8, \frac{5\pi} 8\}$$

Answer 3

$\tan(\theta)= 1$ when $\theta = \frac{(1+4n)\pi}{4}$ where $n$ is an integer starting at $0$, so the zeros for your problem are $2x = \frac{(1+4n)\pi}{4} = \frac{(1+4n)\pi}{8}$. For the range $x$ you gave, that would be the first two values $\frac{\pi}{8},\frac{5\pi}{8}$

Answer 4

root of trigonometric function is not unique

Answer 5

If $f(x)=0$, then $\tan(2x)=1$.

Let $\theta=2x$. For $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$ (the "standard" period of the tangent function), there is precisely one solution to $\tan\theta=1$: namely $\theta=\arctan(1)=\frac{\pi}{4}$. We can find all solutions to $\tan\theta=1$ by taking shifts of this by the period $\pi$: thus the solutions are precisely $$ \theta=\frac{\pi}{4}+k\pi,\qquad k\text{ an integer}. $$ But $\theta=2x$; so, the solutions for $x$ are what we get by dividing both sides by 2: $$ x=\frac{\pi}{8}+\frac{k\pi}{2},\qquad k\text{ an integer}. $$ We must find all specific solutions which lie between $0$ and $3$. These solutions are, then, $$ x=\frac{\pi}{8},\qquad x=\frac{\pi}{8}+\frac{\pi}{2}=\frac{5\pi}{8}. $$ All other solutions are either negative or greater than $3$.

Answer 6

another way to find the roots if to find intersection between f(x) and g(x) f=tan(2x) g=1