Question

\(sin\left(3x+\dfrac{2\pi}{3}\right)+sinx=0\)

Answer 1

\(\Leftrightarrow sin\left(3x+\dfrac{2\pi}{3}\right)=-sinx\)

\(\Leftrightarrow sin\left(3x+\dfrac{2\pi}{3}\right)=sin\left(-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{2\pi}{3}=-x+k2\pi\\3x+\dfrac{2\pi}{3}=\pi+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=-\dfrac{2\pi}{3}+k2\pi\\2x=\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\) (\(k\in Z\))