Question

[Như]

Tìm nghiệm :

2sin(3x+1) = -1 với π<x<2π

Answer 1

[Nguyễn Việt Lâm]

\(\Leftrightarrow sin\left(3x+1\right)=-\frac{1}{2}\)

\(\Rightarrow\left[{}\begin{matrix}3x+1=-\frac{\pi}{6}+k2\pi\\3x+1=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{3}-\frac{\pi}{18}+\frac{k2\pi}{3}\\x=-\frac{1}{3}+\frac{7\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)

Mà \(\pi< x< 2\pi\Rightarrow\left[{}\begin{matrix}\pi< -\frac{1}{3}-\frac{\pi}{18}+\frac{k2\pi}{3}< 2\pi\\\pi>-\frac{1}{3}+\frac{7\pi}{18}+\frac{k2\pi}{3}< 2\pi\end{matrix}\right.\) \(314;628\)

\(\Rightarrow\left[{}\begin{matrix}k=\left\{2;3\right\}\\k=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{3}+\frac{23\pi}{18}\\x=-\frac{1}{3}+\frac{35\pi}{18}\\x=-\frac{1}{3}+\frac{31\pi}{18}\end{matrix}\right.\)