\(\Leftrightarrow sin\left(\dfrac{\pi}{3}cosx-\dfrac{2\pi}{3}-2\pi\right)=0\)
\(\Leftrightarrow sin\left(\dfrac{\pi}{3}cosx-\dfrac{2\pi}{3}\right)=0\)
\(\Leftrightarrow\dfrac{\pi}{3}cosx-\dfrac{2\pi}{3}=k\pi\)
\(\Leftrightarrow cosx=2+3k\) (1)
Do \(-1\le cosx\le1\Rightarrow-1\le2+3k\le1\)
\(\Rightarrow-1\le k\le-\dfrac{1}{3}\Rightarrow k=-1\)
Thế vào (1)
\(\Rightarrow cosx=2-3=-1\)
\(\Leftrightarrow x=\pi+n2\pi\)
\(\Leftrightarrow\dfrac{\pi}{3}\cos x-\dfrac{8\pi}{3}=k\pi\)
\(\Leftrightarrow\dfrac{\pi}{3}\cos x=\dfrac{8\pi}{3}+k\pi\)
\(\Leftrightarrow\cos x=8+3k\)
Mà \(-1\le\cos x\le1\)
\(\Rightarrow-3\le k\le-\dfrac{7}{3}\)
\(\Rightarrow k=-3\)
- Thay k = -3 ta được : \(\cos x=-1\)
\(\Rightarrow x=\pi+k2\pi\)
Vậy ...
