Đáp án:
`sin2x=2sinxcos(x-π/3)`
`<=>sin2x-2sinxsin(x+π/6)=0`
`<=>2sin(π/6-x)sinx=0`
`<=>sin(π/6-x)sinx=0`
`<=>` \(\left[ \begin{array}{l}
\sin(\frac{π}{6}-x)=0\\
\sin x=0
\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}
\frac{π}{6}-x=πn
\\
x = π n
\end{array} \right.\) (với `n\inZ`)
`<=>` \(\left[ \begin{array}{l}
x=\frac{π}{6}-πn
\\
x = π n
\end{array} \right.\) (với `n\inZ`)
Vậy `S={π/6 - π n; π n | n in Z}`
$\\$
`4sin^2x=1`
`<=> sin^2x=1/4`
`<=>` \(\left[ \begin{array}{l}
\sin x = \frac{1}{2}
\\
\sin x = \frac{-1}{2}
\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}
x = \frac{5 π}{6}+2 π n
\\
x = \frac{π}{6}+2 π n
\\
x=\frac{7 π}{6}+2 π n
\\
x=\frac{11 π}{6}+2 π n
\end{array} \right.\) (với `n\inZ`)
Vậy `S={(5 π)/6+2 π n; π/6+2 π n; (7 π)/6+2 π n; (11 π)/6+2 π n | n\in Z}`
$\\$
`2 sin^2(2 x + π/3) = 2 sinx cosx + 1`
`<=> -1 + 2 cos^2(-2 x + π/6) - 2 sinx cosx = 0`
`<=> 2 cos(3 x + π/12) sin(x + π/12) = 0`
`<=> cos(3 x + π/12) sin(x + π/12) = 0`
`<=>` \(\left[ \begin{array}{l}
\cos(3 x + \frac{π}{12}) = 0\\
\sin(x + \frac{π}{12}) = 0
\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}
3 x + \frac{π}{12} = \frac{π}{2}+π n
\\
x + \frac{π}{12} = π n
\end{array} \right.\) (với `n\inZ`)
`<=>` \(\left[ \begin{array}{l}
3 x= \frac{5π}{12}+π n
\\
x = π n - \frac{π}{12}
\end{array} \right.\) (với `n\inZ`)
`<=>` \(\left[ \begin{array}{l}
x= \frac{5π}{36}+\frac{πn}{3}
\\
x = π n - \frac{π}{12}
\end{array} \right.\) (với `n\inZ`)
Vậy `S={(5 π)/36+(π n)/3; π n - π/12 | n in Z}`
