Đáp án:
Giải thích các bước giải:
$\sqrt{3}sin7x-cos7x=2sin(5x-\pi/6)$
⇔ $2sin(7x-\frac{\pi}{6})=2sin(5x-\frac{\pi}{6})$
⇔ $sin(7x-\frac{\pi}{6})=sin(5x-\frac{\pi}{6})$
+) $TH1: 7x-\frac{\pi}{6}=5x-\frac{\pi}{6}+k2\pi$
⇔ $2x=k2\pi$
⇔$x=k\pi$
+) $TH2: 7x-\frac{\pi}{6}=\pi-5x+\frac{\pi}{6}+k2\pi$
⇔ $12x=\frac{4\pi}{3}+k2\pi$
⇔ $x=\frac{\pi}{9}+k\pi/6$
