Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi (k\in Z)$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!
Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!
Đk: $x\neq \frac{\pi}{6}+k\frac{2\pi}{3} \vee -\frac{\pi}{2}+k2\pi
(k\in Z)$Pt<=> $cosx-sin2x=\sqrt{3}(cos2x+sinx)$<=> $sin2x+\sqrt{3}cos2x=cosx-\sqrt{3}sinx$<=> $cos(x-\frac{\pi}{6})=cos(x+\frac{\pi}{3})$Bạn tự tìm nghiệm nhé!