Question

solve for x by factoring where 0 ≤ x ≤ π. tan²x − tan x = 0 ? , π/ , π enter the smallest value first.

Explanation:

Step1: Factor the equation

We have the equation $\tan^{2}x - \tan x = 0$. Factor out $\tan x$ from the left - hand side:
$\tan x(\tan x - 1)=0$

Step2: Solve for $\tan x = 0$

We know that the tangent function $y = \tan x$ has a period of $\pi$, and $\tan x=0$ when $x = n\pi$, where $n$ is an integer. Given the domain $0\leq x\leq\pi$, when $n = 0$, $x = 0$; when $n = 1$, $x=\pi$.

Step3: Solve for $\tan x-1 = 0$

If $\tan x - 1=0$, then $\tan x=1$. The tangent function $\tan x = 1$ when $x=\frac{\pi}{4}+n\pi$, where $n$ is an integer. In the domain $0\leq x\leq\pi$, when $n = 0$, $x=\frac{\pi}{4}$.

Answer:

$0$, $\frac{1}{4}\pi$, $1\pi$ (or in the boxed form as per the problem's requirement: The values are $0$, $\frac{\pi}{4}$, $\pi$)