Chọn C

Chọn B

ĐKXĐ: 

ĐKXĐ:

a. \(cos\left(x-\dfrac{2\pi}{3}\right)\ne0\Rightarrow x-\dfrac{2\pi}{3}\ne\dfrac{\pi}{2}+k\pi\Rightarrow x\ne\dfrac{\pi}{6}+k\pi\)

b. \(sin\left(x+\dfrac{\pi}{6}\right)\ne0\Rightarrow x+\dfrac{\pi}{6}\ne k\pi\Rightarrow x\ne-\dfrac{\pi}{6}+k\pi\)

c. \(\dfrac{1+x}{2-x}\ge0\Rightarrow-1\le x< 2\)

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Hình như đề bài ko đúng đó bn!..bn kiểm tra lại

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

Đặt u = π/2 - x thì u' = -1

Do cos⁡(π/2-x) = sin⁡x