Đáp án:
Giải thích các bước giải:
`\sqrt{3} sin4x - sin5x = cos4x + cos5x`
`⇔ sin (4x-\frac{\pi}{6})=sin (5x+\frac{\pi}{4})`
`⇔` \(\left[ \begin{array}{l}4x-\dfrac{\pi}{6}=5x+\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\\4x-\dfrac{\pi}{6}=\pi-5x-\dfrac{\pi}{4}+k2\pi\ (k \in \mathbb{Z})\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=-\dfrac{5\pi}{12}-k2\pi\ (k \in \mathbb{Z})\\x=\dfrac{11\pi}{108}+k\dfrac{\pi}{9}\ (k \in \mathbb{Z})\end{array} \right.\)
Vậy ......
