Bài 6: Giải các phương trình sau:
a. \(tan (2x + 1)tan (3x - 1) = 1\);
b. \(\tan x + \tan \left( {x + {\pi \over 4}} \right) = 1\)
a) \(tan(2x + 1)tan(3x - 1) = 1\)
\(\tan (2x + 1) = {1 \over {\tan (3x - 1)}}\)
\(\Leftrightarrow \tan (2x + 1) = \cot (3x - 1)\)
\( \Leftrightarrow \tan (2x + 1) = \tan \left( {{\pi \over 2} - 3x + 1} \right)\)
\( \Leftrightarrow 2x + 1 = {\pi \over 2} - 3x + 1 + k\pi \)
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\( \Leftrightarrow x = {\pi \over {10}} + {{k\pi } \over 5}(k \in\mathbb{Z} )\).
b) \(\tan x + \tan \left( {x + {\pi \over 4}} \right) = 1\)
\(\eqalign{
& \Leftrightarrow \tan x + {{\tan x + \tan {\pi \over 4}} \over {1 - \tan x.\tan {\pi \over 4}}} = 1 \cr
& \Leftrightarrow \tan x + {{\tan x + 1} \over {1 - \tan x}} = 1 \cr} \)
Đặt \(t = tan x\), (điều kiện \(t ≠ 1\))phương trình trở thành
\(t + \frac{t+1}{1-t}\)= 1
\(\Leftrightarrow - {t^2} + 3t = 0 \Leftrightarrow \left[ \matrix{
t = 0 \hfill \cr
t = 3 \hfill \cr} \right.\text{(thỏa mãn)}\)
\( \Leftrightarrow \left[ \matrix{
\tan x = 0 \hfill \cr
\tan x = 3 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x = k\pi \hfill \cr
x = \arctan 3 + k\pi \hfill \cr} \right.(k \in \mathbb{Z})\)