Tìm nghiệm phương trình

Đáp án:

Giải thích các bước giải:

 $\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=-sin6x$

$cos\dfrac{\pi}{3}sin2x+cos2x.sin\dfrac{\pi}{3}=-sin6x$
$sin(2x+\dfrac{\pi}{3})=sin(-6x)$

\(\left[ \begin{array}{l}2x+\dfrac{\pi}{3}=-6x+k2\pi\\2x+\dfrac{\pi}{3}=\pi+6x+k2\pi\end{array} \right.\) 

\(\left[ \begin{array}{l}8x=-\dfrac{\pi}{3}+k2\pi\\-4x=\dfrac{2\pi}{3}+k2\pi\end{array} \right.\)

\(\left[ \begin{array}{l}x=-\dfrac{\pi}{24}+k\dfrac{\pi}{4}\\x=\dfrac{-\pi}{6}+k\dfrac{\pi}{2}\end{array} \right., k\in Z\) 

Vậy $S=\{\ -\dfrac{\pi}{24}+k\dfrac{\pi}{4} ; \dfrac{-\pi}{6}+k\dfrac{\pi}{2} |k\in Z \}$