Đáp án:
$x=\dfrac{\pi}4+k\pi$ và $x=k\pi$ $(k\in\mathbb Z)$
Lời giải:
\({\tan}^3\left({x-\dfrac{\pi}{4}}\right)=\tan x-1\)
ĐKXĐ:
\(\left\{ \begin{array}{l} \cos\left({x-\dfrac{\pi}{4}}\right)\ne0 \\ \cos x\ne0 \end{array} \right .\Leftrightarrow \left\{ \begin{array}{l} x-\dfrac{\pi}{4}\ne \dfrac{\pi}{2}+k\pi \\ x\ne \dfrac{\pi}{2}+k\pi \end{array} \right .\)
\(\Leftrightarrow \left\{ \begin{array}{l} x\ne\dfrac{3\pi}{4}+k\pi \\ x\ne\dfrac{\pi}{2}+k\pi \end{array} \right .\) (\(k\in\mathbb Z\))
PT \(\Rightarrow \left[{\dfrac{\tan x-\tan\dfrac{\pi}{4}}{1+\tan x\tan\dfrac{\pi}{4}}}\right]^3=\tan x-1\)
\(\Rightarrow \left({\dfrac{\tan x-1}{1+\tan x}}\right)^3=\tan x-1\)
\(\Rightarrow \dfrac{(\tan x-1)^3}{(1+\tan x)^3}-(\tan x-1)=0\)
\(\Rightarrow (\tan x-1)\left[{\dfrac{(\tan x-1)^2}{(1+\tan x)^3}-1}\right]=0\)
\(\Rightarrow \left[ \begin{array}{l} \tan x-1=0\text{ (1)} \\ \dfrac{(\tan x-1)^2}{(1+\tan x)^3}-1=0\text{ (2)} \end{array} \right .\)
(1) \( \Leftrightarrow \tan x=1\Rightarrow x=\dfrac{\pi}{4}+k\pi,(k\in\mathbb Z)\) (tm)
(2) \(\Rightarrow (\tan x-1)^2-(1+\tan x)^3=0\)
\(\Rightarrow {\tan}^3x+2{\tan}^2x+5\tan x=0\)
\(\Rightarrow \tan x=0\)
\(\Rightarrow x=k\pi,(k\in\mathbb Z)\) (tm).
