Question

\(sin\left(\dfrac{\pi}{3}-x\right)=cos\left(\dfrac{2\pi}{3}+x\right)\)

Answer 1

Đặt \(t=\dfrac{\pi}{3}-x\)

\(\Rightarrow\pi-t=\dfrac{2\pi}{3}+x\)

Từ đó phương trình đã cho trở thành:

\(sint=cos\left(\pi-t\right)\\ \Leftrightarrow sint=-cost\\ \Leftrightarrow sint=cos\left(\pi+t\right)\\ \Leftrightarrow sint=sin\left(\dfrac{\pi}{2}-\pi-t\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{\pi}{2}-t+k2\pi\\t=\dfrac{3\pi}{2}+t+k2\pi\left(VN\right)\end{matrix}\right.\)

\(\Leftrightarrow t=-\dfrac{\pi}{4}+k\pi\\ \Leftrightarrow\dfrac{\pi}{3}-x=-\dfrac{\pi}{4}+k\pi\\ \Leftrightarrow x=\dfrac{7\pi}{12}+k\pi\)

\(\left(k\in Z\right)\)