Question

\(\sin\left(x-\dfrac{\pi}{4}\right)=-sin\left(2x-\dfrac{\pi}{6}\right)\)

Answer 1

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=-sin\left(x-\dfrac{\pi}{4}\right)\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{4}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{6}=\dfrac{\pi}{4}-x+k2\pi\\2x-\dfrac{\pi}{6}=\dfrac{3\pi}{4}+x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{36}+\dfrac{k2\pi}{3}\\x=\dfrac{11\pi}{12}+k2\pi\end{matrix}\right.\) (\(k\in Z\))