Đáp án:
$a)
{\left[\begin{aligned}k=0\\ k=1\end{aligned}\right.}\\$
b)
Không có k nào thỏa mãn
Giải thích các bước giải:
$a)
\sin2x=\frac{-1}{2}\\
\Leftrightarrow \sin2x=\sin\frac{-\pi}{6}\\
\Leftrightarrow {\left[\begin{aligned}2x=\frac{-\pi}{6}+k2\pi\\2x=\pi+\frac{\pi}{6}+k2\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\frac{-\pi}{12}+k\pi\\2x=\frac{7\pi}{6}+k2\pi\end{aligned}\right.}\Leftrightarrow {\left[\begin{aligned}x=\frac{-\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{aligned}\right.}\\
0<x<\pi\Rightarrow {\left[\begin{aligned}0<\frac{-\pi}{12}+k\pi<\pi\\ 0<\frac{7\pi}{12}+k\pi<\pi\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\frac{\pi}{12}<k\pi<\pi+\frac{\pi}{12}\\ \frac{-7\pi}{12}<k\pi<\pi-\frac{7\pi}{12}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\frac{\pi}{12}<k\pi<\frac{13\pi}{12}\\ \frac{-7\pi}{12}<k\pi<\frac{5\pi}{12}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\frac{1}{12}<k<\frac{13}{12}\\ \frac{-7}{12}<k<\frac{5}{12}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}k=0\\ k=1\end{aligned}\right.}\\
b)
ĐK: \sin3x\neq 0\Leftrightarrow x\neq \frac{k2\pi}{3}\\
\cot3x=\frac{-1}{\sqrt{3}}\\
\Leftrightarrow \cot3x=\cot\frac{2\pi}{3}\\
\Leftrightarrow 3x=\frac{2\pi}{3}+k\pi\\
\Leftrightarrow x=\frac{2\pi}{9}+\frac{k\pi}{3}\\
\frac{-\pi}{2}<x<0\Rightarrow \frac{-\pi}{2}<\frac{2\pi}{9}+\frac{k\pi}{3}<0\\
\Rightarrow \frac{-\pi}{2}-\frac{2\pi}{9}<\frac{k\pi}{3}<-\frac{2\pi}{9}\\
\Rightarrow \frac{-13\pi}{18}<\frac{k\pi}{3}<-\frac{2\pi}{9}\\
\Rightarrow \frac{-13}{18}<\frac{k}{3}<-\frac{2}{9}\\
\Rightarrow \frac{-13}{6}<k<-\frac{2}{3}$
Không có k nào thỏa mãn
