Đáp án:
Giải thích các bước giải:
`\sqrt{3} cos 5x-2sin3xcos2x-sinx=0`
`⇔ \sqrt{3} cos 5x-(sin 5x+sin x)-sin x=0`
`⇔ \sqrt{3} cos 5x-sin 5x=2sin x`
`⇔ sin (\frac{\pi}{3}-5x)=sin x`
`⇔` \(\left[ \begin{array}{l}\dfrac{\pi}{3}-5x=x+k2\pi\ (k \in \mathbb{Z})\\\dfrac{\pi}{3}-5x=\pi-x+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=\dfrac{\pi}{18}-k\dfrac{\pi}{3}\ (k \in \mathbb{Z})\\x=-\dfrac{\pi}{6}-k\dfrac{\pi}{2}\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy .............
