Đáp án:
`x\in {π/3;π/2;{4π}/3;{3π}/2;{7π}/3;{5π}/2}`
Giải thích các bước giải:
`\qquad sin(2x-π/3)=\sqrt{3}/2=sin \ π/3`
`=>`$\left[\begin{array}{l}2x-\dfrac{π}{3}=\dfrac{π}{3}+k2π\\2x-\dfrac{π}{3}=π-\dfrac{π}{3}+k2π\end{array}\right.$
`=>`$\left[\begin{array}{l}2x=\dfrac{2π}{3}+k2π\\2x=π+k2π\end{array}\right.$
`=>`$\left[\begin{array}{l}x=\dfrac{π}{3}+kπ\\x=\dfrac{π}{2}+kπ\end{array}\right.$`(k\in ZZ)`
Vì `x\in (0;3π)`
+) `0<x=π/3+kπ<3π` `\quad (k\in ZZ)`
`=>x\in {π/3; π/3+π; π/3+2π}`
`=>x\in {π/3;{4π}/3; {7π}/3}`
$\\$
+) `0<x=π/2+kπ<3π` `\quad (k\in ZZ)`
`=>x\in {π/2; π/2+π; π/2+2π}`
`=>x\in {π/2;{3π}/2; {5π}/2}`
$\\$
Vậy nghiệm của phương trình thuộc khoảng `(0;3π)` là:
`x\in {π/3;π/2;{4π}/3;{3π}/2;{7π}/3;{5π}/2}`
