Đáp án:
`sin2x=(-1)/(2)=(-π)/(6)`
`<=>`\(\left[ \begin{array}{l}2x=\dfrac{-π}{6}+k2π\\2x=π-\dfrac{-π}{6}+k2π\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=\dfrac{-π}{6}+k2π\\2x=\dfrac{6π}{6}+\dfrac{π}{6}+k2π\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}2x=\dfrac{-π}{6}+k2π\\2x=\dfrac{7π}{6}+k2π\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=(\dfrac{-π}{6}+k2π):2\\x=(\dfrac{7π}{6}+k2π):2\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{-π}{12}+kπ\\x=\dfrac{7π}{12}+kπ\end{array} \right.\) `(k∈Z)`
