\(\Leftrightarrow x-\dfrac{\pi}{3}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{5\pi}{6}+k2\pi\left(k\in Z\right)\)
`sin (x-\pi/3)=1`
`<=>x-\pi/3=\pi/2+k2\pi` , `k in ZZ`
`<=>x=[5\pi]/6 +k2\pi` , `k in ZZ`
Vậy ptr có nghiệm `x=[5\pi]/6 +k2\pi` , `k in ZZ`