`~rai~`
\(\text{Câu 16:}\\\sin\left(3x+\dfrac{\pi}{4}\right)=\sin x\\\Leftrightarrow \left[\begin{array}{I}3x+\dfrac{\pi}{4}=x+k2\pi\\3x+\dfrac{\pi}{4}=\pi-x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}2x=-\dfrac{\pi}{4}+k2\pi\\4x=\dfrac{3\pi}{4}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{\pi}{8}+k\pi\\x=\dfrac{3\pi}{16}+k\dfrac{\pi}{2}.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{-\dfrac{\pi}{8}+k\pi;\dfrac{3\pi}{16}+k\dfrac{\pi}{2}\Big|k\in\mathbb{Z}\right\}\\\to\text{Chọn ý B.}\\\text{Câu 17:}\\\cos(2x+1)=\cos(x-3)\\\Leftrightarrow \left[\begin{array}{I}2x+1=x-3+k2\pi\\2x+1=3-x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-4+k2\pi\\3x=2+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-4+k2\pi\\x=\dfrac{2}{3}+k\dfrac{2\pi}{3}.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{-4+k2\pi;\dfrac{2}{3}+k\dfrac{2\pi}{3}\Big|k\in\mathbb{Z}\right\}.\\\to\text{Chọn ý C.}\\\text{Câu 18:}\\\tan5x-\tan x=0\quad(1)\\ĐKXĐ:\begin{cases}\cos x\ne 0\\\cos 5x\ne 0\end{cases}\\\Leftrightarrow \begin{cases}x\ne\dfrac{\pi}{2}+m\pi\\x\ne\dfrac{\pi}{10}+m\dfrac{\pi}{5}\end{cases}\quad(m\in\mathbb{Z})\\(1)\Leftrightarrow \tan5x=\tan x\\\Leftrightarrow 5x=x+k\pi\\\Leftrightarrow 4x=k\pi\\\Leftrightarrow x=k\dfrac{\pi}{4}.(k\in\mathbb{Z})\\\text{Kết hợp với ĐKXĐ:}\\\begin{cases}k\dfrac{\pi}{4}\ne\dfrac{\pi}{10}+m\dfrac{\pi}{5}\\k\dfrac{\pi}{4}\ne\dfrac{\pi}{2}+m\pi\end{cases}\\\Leftrightarrow k\dfrac{\pi}{4}\ne \dfrac{\pi}{2}+m\pi\\\Leftrightarrow \dfrac{1}{4}k\ne\dfrac{1}{2}+m\\\Leftrightarrow k\ne 4m+2.\\\text{Vậy S=}\left\{k\dfrac{\pi}{4}\Big|k\in\mathbb{Z};k\ne4m+2;m\in\mathbb{Z}\right\}.\\\to\text{Chọn ý C.}\\\text{Câu 19:}\\\cos(2x+60^\circ)=\cos x\\\Leftrightarrow \left[\begin{array}{I}2x+60^\circ=x+k360^\circ\\2x+60^\circ=-x+k360^\circ\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-60^\circ+k360^\circ\\3x=-60^\circ+k360^\circ\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-60^\circ+k360^\circ\\x=-20^\circ+k120^\circ.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\{-60^\circ+k360^\circ;-20^\circ+k120^\circ|k\in\mathbb{Z}\}.\\\to\text{Chọn ý A.}\\\text{Câu 20:}\\\sin\left(3x+\dfrac{\pi}{3}\right)=\cos x\\\Leftrightarrow \sin\left(3x+\dfrac{\pi}{3}\right)=\sin\left(\dfrac{\pi}{2}-x\right)\\\Leftrightarrow \left[\begin{array}{I}3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}-x+k2\pi\\3x+\dfrac{\pi}{3}=\pi-\dfrac{\pi}{2}+x+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}4x=\dfrac{\pi}{6}+k2\pi\\2x=\dfrac{\pi}{6}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{\pi}{24}+k\dfrac{\pi}{2}\\x=\dfrac{\pi}{12}+k\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy S=}\left\{\dfrac{\pi}{24}+k\dfrac{\pi}{2};\dfrac{\pi}{12}+k\pi\Big|k\in\mathbb{Z}\right\}.\\\to\text{Chọn ý A.}\)
