\(sin\left(x+\frac{\pi}{3}\right)-cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=cosx\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}\right)=sin\left(\frac{\pi}{2}-x-\frac{\pi}{3}\right)\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{3}\right)=sin\left(\frac{\pi}{6}-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{3}=\frac{\pi}{6}-x+k2\pi,k\in Z\\x+\frac{\pi}{3}=\pi-\frac{\pi}{6}+x+k2\pi,k\in Z\end{matrix}\right.\)
\(\Leftrightarrow2x=-\frac{\pi}{6}+k2\pi,k\in Z\)
\(\Leftrightarrow x=-\frac{\pi}{12}+k\pi,k\in Z\)
Vậy phương trình trên có nghiệm là \(S=\left\{-\frac{\pi}{12}+k\pi,k\in Z\right\}\)
