\(\Leftrightarrow2x=x-\dfrac{\pi}{4}+k\pi\)

\(\Leftrightarrow x=-\dfrac{\pi}{4}+k\pi\) (\(k\in Z\))

Em nghĩ là sai. ĐKXĐ là x ≠ \(\dfrac{\pi}{4}\) + k . \(\dfrac{\pi}{2}\)

Phương trình vô nghiệm 

Điều kiện:

⇔ tan x.(1 - tanx) + tanx + 1 = 1 – tan x.

⇔ tan x - tan2x + 2.tan x = 0

⇔ tan2x - 3tanx = 0

⇔ tanx(tanx - 3) = 0

Vậy phương trình đã cho có tập nghiệm là:

{arctan 3+kπ; k ∈ Z }

Chọn A

Câu 1: \(\tan x=\tan\left(\frac{6\pi}{5}\right)\)

=>\(x=\frac{6\pi}{5}+k\pi\)

=>Nghiệm nguyên dương nhỏ nhất là \(\frac{6\pi}{5}-\pi=\frac15\pi\)

=>Chọn A

Câu 2: \(\cot2x=\cot\left(\frac{\pi}{2}-x\right)\)

=>\(2x=\frac{\pi}{2}-x+k\pi\)

=>\(3x=\frac{\pi}{2}+k\pi\)

=>\(x=\frac{\pi}{6}+\frac{k\pi}{3}\)

mà \(x\in\left\lbrack0;\pi\right\rbrack\)

nên \(x\in\left\lbrace\frac{\pi}{6};\frac{\pi}{2};\frac56\pi\right\rbrace\)

=>Chọn B

Câu 3:

\(4\cdot sin^22x-1=0\)

=>\(4\cdot sin^22x=1\)

=>\(\sin^22x=\frac14\)

=>\(\left[\begin{array}{l}\sin2x=\frac12\\ \sin2x=-\frac12\end{array}\right.\)

TH1: sin 2x=1/2

=>\(\left[\begin{array}{l}2x=\frac{\pi}{6}+k2\pi\\ 2x=\pi-\frac{\pi}{6}+k2\pi=\frac56\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{12}+k\pi\\ x=\frac{5}{12}\pi+k\pi\end{array}\right.\)

mà \(x\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\)

nên \(x\in\left(\frac{\pi}{12};\frac{5}{12}\pi;-\frac{1}{12}\pi\right)\)

TH2: sin 2x=-1/2

=>\(\left[\begin{array}{l}2x=\frac{-\pi}{6}+k2\pi\\ 2x=\pi-\frac{-\pi}{6}+k2\pi=\frac76\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac{\pi}{12}+k\pi\\ x=\frac{7}{12}\pi+k\pi\end{array}\right.\)

mà \(x\in\left(-\frac{\pi}{2};\frac{\pi}{2}\right)\)

nên \(x\in\left(-\frac{\pi}{12};-\frac{5}{12}\pi\right)\)

Tổng các nghiệm là \(\frac{\pi}{12}+\frac{5\pi}{12}-\frac{1}{12}\pi-\frac{\pi}{12}-\frac{5}{12}\pi=-\frac{1}{12}\pi\)

Câu 4: \(cos\left(x+\frac{\pi}{4}\right)=\frac12\)

=>\(\left[\begin{array}{l}x+\frac{\pi}{4}=\frac{\pi}{3}+k2\pi\\ x+\frac{\pi}{4}=-\frac{\pi}{3}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{12}+k2\pi\\ x=-\frac{7}{12}\pi+k2\pi\end{array}\right.\)

mà \(x\in\left(-\pi;\pi\right)\)

nên \(x\in\left(\frac{\pi}{12};-\frac{7}{12}\pi\right)\)

=>Tổng các nghiệm là:

\(\frac{\pi}{12}-\frac{7}{12}\pi=-\frac{6}{12}\pi=-\frac12\pi\)

=>Chọn B

ĐK: \(x\ne\dfrac{5\pi}{12}+\dfrac{k\pi}{2}\)

\(tan\left(2x-\dfrac{\pi}{3}\right)=-\dfrac{1}{2}\)

\(\Leftrightarrow2x-\dfrac{\pi}{3}=arctan\left(-\dfrac{1}{2}\right)+k\pi\)

\(\Leftrightarrow2x=\dfrac{\pi}{3}+arctan\left(-\dfrac{1}{2}\right)+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{1}{2}arctan\left(-\dfrac{1}{2}\right)+\dfrac{k\pi}{2}\in\left(0;\pi\right)\)

...

\(sin\left(x-\dfrac{\pi}{2}\right)+cos\left(x-\pi\right)+tan\left(\dfrac{5\pi}{2}-x\right)+tan\left(x-\dfrac{\pi}{2}\right)\)

\(=-sin\left(\dfrac{\pi}{2}-x\right)+cos\left(\pi-x\right)+tan\left(2\pi+\dfrac{\pi}{2}-x\right)-tan\left(\dfrac{\pi}{2}-x\right)\)

\(=-cosx-cosx+tan\left(\dfrac{\pi}{2}-x\right)-cotx\)

\(=-2cosx+cotx-cotx=-2cosx\)

Lê Huy Hoàng:

a) ĐK: $x\in\mathbb{R}\setminus \left\{k\pi\right\}$ với $k$ nguyên

PT $\Leftrightarrow \tan ^2x-4\tan x+5=0$

$\Leftrightarrow (\tan x-2)^2+1=0$

$\Leftrightarrow (\tan x-2)^2=-1< 0$ (vô lý)

Do đó pt vô nghiệm.

c)

ĐK:.............

PT $\Leftrightarrow 1+\frac{\sin ^2x}{\cos ^2x}-1+\tan x-\sqrt{3}(\tan x+1)=0$

$\Leftrightarrow \tan ^2x+\tan x-\sqrt{3}(\tan x+1)=0$

$\Leftrightarrow \tan ^2x+(1-\sqrt{3})\tan x-\sqrt{3}=0$

$\Rightarrow \tan x=\sqrt{3}$ hoặc $\tan x=-1$

$\Rightarrow x=\pi (k-\frac{1}{4})$ hoặc $x=\pi (k+\frac{1}{3})$ với $k$ nguyên

d)

ĐK:.......

PT $\Leftrightarrow \tan x-\frac{2}{\tan x}+1=0$

$\Leftrightarrow \tan ^2x+\tan x-2=0$

$\Leftrightarrow (\tan x-1)(\tan x+2)=0$

$\Rightarrow \tan x=1$ hoặc $\tan x=-2$

$\Rightarrow x=k\pi +\frac{\pi}{4}$ hoặc $x=k\pi +\tan ^{-2}(-2)$ với $k$ nguyên.