Đáp án:
Tham khảo
Giải thích các bước giải:
$\sqrt{3}cos5x-2sin3xcos2x-sinx=0$
$⇔\sqrt{3}cos5x-5sinx=2sĩn$
$⇔\dfrac{\sqrt{3}}{2}cos5x-\dfrac{1}{2}sin5x=sĩn$
$⇔sin(\dfrac{π}{3}-5x)=sinx$
⇔\(\left[ \begin{array}{l}\dfrac{π}{3}-5x=x+k2π\\\dfrac{π}{3}-5x=π-x+k2π\end{array} \right.\)
⇔\(\left[ \begin{array}{l}6x\dfrac{π}{3}-k2π\\4x=\dfrac{π}{3}-π-k2π=-\dfrac{2π}{3}-k2π\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x=\dfrac{π}{18}-k\dfrac{π}{3}\\x=-\dfrac{π}{6}-k\dfrac{π}{2}\end{array} \right.\) $(k∈\mathbb{Z})$
