Question

Giúp mình giải mẫu sin(x-120°)+ cos2x=0

Answer 1

Lời giải:
\(\sin (x-120^0)+\cos 2x=0\)

\(\sin (x-\frac{2\pi}{3})=-\cos 2x=\cos (\pi-2x)=\sin [\frac{\pi}{2}-(\pi-2x)]\)

\(=\sin (2x+\frac{\pi}{2})\)

\(\Rightarrow \left[\begin{matrix} x-\frac{2\pi}{3}=2x+\frac{\pi}{2}+2k\pi\\ x-\frac{2\pi}{3}=\pi -(2x+\frac{\pi}{2})+2k\pi\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=\frac{-7}{6}\pi -2k\pi\\ x=\frac{\pi}{18}+\frac{2k+1}{3}\pi\end{matrix}\right.\) với $k$ nguyên bất kỳ.