Giải các phương trình sau:

a) tan(2x+1)tan(2x-1)=1;                   b)tanx+tan$\left(x+\frac{\mathrm{\pi }}{4}\right)=1$

Giải:

a) Điều kiện cos(2x+1)≠0, cos(3x-1)≠0

tan(2x+1)tan(3x-1)=0

$$\begin{gathered} \iff \sin (2x+1)\sin (3x-1)=\cos (2x+1)\cos (3x-1) \\ \iff \cos (2x+1)\cos (3x-1)-\sin (2x+1)\sin (3x-1)=0 \\ \iff \cos \left[\right(2x+1+(3x-1)]=0\iff \cos 5x=0 \\ \iff 5x=\frac{\mathrm{\pi }}{2}+k\mathrm{\pi } \\ \iff \mathrm{x}=\frac{\mathrm{\pi }}{10}+\mathrm{k}\frac{\mathrm{\pi }}{5}, \mathrm{k}\in \mathrm{\mathbb{Z}} \end{gathered}$$

b) Điều kiện cosx≠0; $\cos (x-\frac{\mathrm{\pi }}{4})\ne 0$

tanx +$\tan (x+\frac{\mathrm{\pi }}{4})=1 \iff \tan x+\frac{\tan x+1}{1-\tan x}=1$

$\iff \tan x-{\tan }^{2}x+\tan x+1=1-\tan x\iff {\tan }^{2}x-3\tan x=0$

tanx(tanx - 3)=0$\iff {[}_{\tan x=3}^{\tan x=0}\iff {[}_{x=arc\tan 3+k\mathrm{\pi }}^{x=k\mathrm{\pi }} (k\in \mathrm{\mathbb{Z}})$