Question

Answer 1

Ta có: \(\sin x-\sin3x-\sqrt3\left(cosx+cos3x\right)=0\)

=>\(sinx-\sqrt3\cdot cosx=\sin3x+\sqrt3\cdot cos3x\)

=>\(\frac12\cdot\sin x-\frac{\sqrt3}{2}\cdot cosx=\frac12\cdot\sin3x+\frac{\sqrt3}{2}\cdot cos3x\)

=>\(\sin\left(3x+\frac{\pi}{3}\right)=\sin\left(x-\frac{\pi}{3}\right)\)

=>\(\left[\begin{array}{l}3x+\frac{\pi}{3}=x-\frac{\pi}{3}+k2\pi\\ 3x+\frac{\pi}{3}=\pi-x+\frac{\pi}{3}+k2\pi=\frac43\pi-x+k2\pi\end{array}\right.\)

=>\(\left[\begin{array}{l}2x=-\frac23\pi+k2\pi\\ 4x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\pi+k\pi\\ x=\frac{\pi}{4}+\frac{k\pi}{2}\end{array}\right.\)