Đáp án:
\(\left[ \begin{array}{l}x=\frac{-\pi}6+k2\pi\\x=\frac{11\pi}{8}+\frac{k2\pi}{3}\end{array} \right.\) $(k∈Z)$
Giải thích các bước giải:
$sin(2x-$$\frac{\pi}{3})=-cosx$
⇔$sin(2x-\frac{\pi}{3})=cos(\pi-x)$
⇔$sin(2x-\frac{\pi}{3})=sin(x-\frac{\pi}2)$
⇔\(\left[ \begin{array}{l}2x-\frac{\pi}3=x-\frac\pi2+k2\pi\\2x-\frac{\pi}3=\pi-x+\frac{\pi}2+k2\pi\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=\frac{-\pi}6+k2\pi\\x=\frac{11\pi}{8}+\frac{k2\pi}{3}\end{array} \right.\) $(k∈Z)$
