`sin (3x+20^o)=-1/2`
`<=>` $\left[\begin{matrix} 3x+20^o =-30^o +k360^o\\ 3x+20^o =210^o +k360^o\end{matrix}\right.$
`<=>` $\left[\begin{matrix} x=\dfrac{-50^o }{3}+k120^o \\ x=\dfrac{190^o }{3}+k120^o\end{matrix}\right.$ `(k in ZZ)`
Ta có :
\(sin\left(3x+\dfrac{\pi}{9}\right)=\dfrac{-\pi}{6}\)
⇔ \(\left[{}\begin{matrix}3x+\dfrac{\pi}{9}=\dfrac{-\pi}{6}+k2\pi\\3x+\dfrac{\pi}{9}=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}3x=\dfrac{-5\pi}{18}+k2\pi\\3x=\dfrac{19\pi}{18}+k2\pi\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=-\dfrac{5\pi}{54}+\dfrac{k2\pi}{3}\\x=\dfrac{19\pi}{54}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
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