ĐKXĐ: 1-2*sin x<>0
=>\(2\cdot\sin x<>1\)
=>\(\sin x<>\frac12\)
=>\(x\notin\left\lbrace\frac{\pi}{6}+k2\pi;\frac56\pi\,+k2\pi\right\rbrace\)
\(\frac{cos\left(\frac{\pi}{2}-2x\right)-\sqrt3\cdot cos\left(\pi-2x\right)-2\cdot cosx}{1-2\cdot\sin x}=0\)
=>\(cos\left(\frac{\pi}{2}-2x\right)-\sqrt3\cdot cos\left(\pi-2x\right)-2\cdot cosx=0\)
=>sin 2x-\(\sqrt3\cdot\left(-cos2x\right)-2\cdot cosx=0\)
=>\(\sin2x+\sqrt3\cdot cos2x=2\cdot cosx\)
=>\(\frac12\cdot\sin2x+\frac{\sqrt3}{2}\cdot cos2x=cosx\)
=>\(\sin\left(2x+\frac{\pi}{3}\right)=\sin\left(\frac{\pi}{2}-x\right)\)
=>\(\left[\begin{array}{l}2x+\frac{\pi}{3}=\frac{\pi}{2}-x+k2\pi\\ 2x+\frac{\pi}{3}=\pi-\frac{\pi}{2}+x+k2\pi=x+\frac{\pi}{2}+k2\pi\end{array}\right.\)
=>\(\left[\begin{array}{l}2x+x=\frac{\pi}{2}-\frac{\pi}{3}+k2\pi=\frac{\pi}{6}+k2\pi\\ 2x-x=\frac{\pi}{2}-\frac{\pi}{3}+k2\pi=\frac{\pi}{6}+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{\pi}{18}+\frac{k2\pi}{3}\\ x=\frac{\pi}{6}+k2\pi\end{array}\right.\)
=>\(x=\frac{\pi}{18}+\frac{k2\pi}{3}\)
mà \(x\notin\left\lbrace\frac{\pi}{6}+k2\pi;\frac56\pi\,+k2\pi\right\rbrace\)
nên \(x\in\left\lbrace\frac{\pi}{18}+k2\pi;\frac{13}{18}\pi+k2\pi;\frac{25}{18}\pi+k2\pi\right\rbrace\)
