8 câu Dạng 3: Phương trình tanx=m

Điều kiện $\cos \left(5x+\frac{\pi }{4}\right)\ne 0\iff 5x+\frac{\pi }{4}\ne \frac{\pi }{2}+k\pi \iff x\ne \frac{\pi }{20}+\frac{k\pi }{5}$  , $\left(k\in \mathbb{Z}\right)$ .

$\left(1\right)\iff \tan \left(5x+\frac{\pi }{4}\right)=\frac{\sqrt{3}}{3}\iff \tan \left(5x+\frac{\pi }{4}\right)=\tan \frac{\pi }{6}$, 

$\iff 5x+\frac{\pi }{4}=\frac{\pi }{6}+k\pi \iff 5x=-\frac{\pi }{12}+k\pi \iff x=-\frac{\pi }{60}+k\frac{\pi }{5}$

Vậy phương trình đã cho có nghiệm là $x=-\frac{\pi }{60}+k\frac{\pi }{5}$ , $\left(k\in \mathbb{Z}\right)$.

Điều kiện $\left\{\begin{array}{l}\cos \left(2x-\frac{\pi }{4}\right)\ne 0\\ \sin x\ne 0\end{array}\right.\iff \left\{\begin{array}{l}2x-\frac{\pi }{4}\ne \frac{\pi }{2}+k\pi \\ x\ne l\pi \end{array}\right.\iff \left\{\begin{array}{l}x\ne \frac{3\pi }{8}+\frac{k\pi }{2}\\ x\ne l\pi \end{array}\right.$   .

$\left(2\right)\iff \tan \left(2x-\frac{\pi }{4}\right)=\tan \left(\frac{\pi }{2}-x\right)\iff 2x-\frac{\pi }{4}=\frac{\pi }{2}-x+k\pi \iff x=\frac{\pi }{4}+\frac{k\pi }{3}$, $\left(k\in \mathbb{Z}\right)$ .

Vậy phương trình đã cho có nghiệm là $x=\frac{\pi }{4}+\frac{k\pi }{3},(k\in \mathbb{Z})$  .

Ta có $\tan 45°=1\iff \tan \left(x+15°\right)=\tan 45°\iff x+15°=45°+k.180°\iff x=30°+k.180°$ .

Với $90°<x<270°\iff 90°<30°+k.180°<270°\Rightarrow k=1\Rightarrow x=210°$ .

Đáp án D

Phương trình $\sqrt{3}.\tan x+3=0$  có nghĩa $\iff \cos x\ne 0\iff x\ne \frac{\pi }{2}+k\pi \iff D=ℝ\backslash \left\{\frac{\pi }{2}+k\pi \right\}$ .

Ta có $\sqrt{3}\tan x+3=0\iff \tan x=-\sqrt{3}\iff \tan x=\tan \frac{-\pi }{3}\iff x=-\frac{\pi }{3}+k\pi$  $\left(k\in \mathbb{Z}\right)$ .

Phương trình ${\tan }^{2}x=3$  có nghĩa $\iff \cos x\ne 0\iff x\ne \frac{\pi }{2}+k\pi \iff D=ℝ\backslash \left\{\frac{\pi }{2}+k\pi \right\}$ .

Ta có ${\tan }^{2}x=3\iff \left[\begin{array}{l}\tan x=\sqrt{3}\\ \tan x=-\sqrt{3}\end{array}\right.$ .

Xét $\tan x=\sqrt{3}\iff \tan x=\tan \frac{\pi }{3}\iff x=\frac{\pi }{3}+k\pi$ .

Xét $\tan x=-\sqrt{3}\iff \tan x=\tan \frac{-\pi }{3}\iff x=\frac{-\pi }{3}+k\pi$ .

Vậy $x=\pm \frac{\pi }{3}+k\pi$ ,$\left(k\in \mathbb{Z}\right)$ .

Phương trình $\tan x=-\tan \frac{\pi }{5}$  có nghĩa $\iff \cos x\ne 0\iff x\ne \frac{\pi }{2}+k\pi \iff D=ℝ\backslash \left\{\frac{\pi }{2}+k\pi \right\}$ .

Ta có $\tan x=-\tan \frac{\pi }{5}\iff \tan x=\tan \frac{-\pi }{5}\iff x=\frac{-\pi }{5}+k\pi$ .

Do $x\in \left(\frac{\pi }{2};\pi \right)$  nên $x=\frac{4\pi }{5}$ .