Question

 

\(\sin\left(3x+20^o\right)=-\dfrac{1}{2}\)

Answer 1

`sin 2x=[-\sqrt{2}]/2`

`<=>` $\left[\begin{matrix} 2x=\dfrac{-\pi}{4}+k2\pi\\ 2x=\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.$

`<=>` $\left[\begin{matrix} x=\dfrac{-\pi}{8}+k\pi\\ x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.$   `(k in ZZ)`

Answer 2

sin 2x \(=\dfrac{-\sqrt{2}}{2}\Leftrightarrow sin2x=sin\dfrac{-\pi}{4}\)   \(\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{\pi}{4}+2k\pi\\2x=\dfrac{5\pi}{4}+2k\pi\end{matrix}\right.\)  ( k thuộc Z )

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\pi}{8}+k\pi\\x=\dfrac{5\pi}{8}+k\pi\end{matrix}\right.\)