Đáp án:
`cos3 x + sin2 x = 0`
`<=> cos3 x = - sin2 x`
`<=> cos(3 x) = cos(-2 x - π/2)`
`<=>` \(\left[ \begin{array}{l}
3x = \frac{3 π}{2} - 2 x + 2πn
\\
3x = \frac{-3 π}{2} + 2 x + 2 π n
\end{array} \right.\) (với `n\inZ`)
`<=>` \(\left[ \begin{array}{l}
5x = \frac{3 π}{2} + 2πn
\\
x = \frac{-3 π}{2} + 2 π n
\end{array} \right.\) (với `n\inZ`)
`<=>` \(\left[ \begin{array}{l}
x = \frac{3 π}{10} + \frac{2πn}{5}
\\
x = \frac{-3 π}{2} + 2 π n
\end{array} \right.\) (với `n\inZ`)
Vậy `S={(3 π)/10+(2 π n)/5; - (3 π)/2+2 π n | n \inZ}`
$\\$
`-------------------`
`sin(2x-pi/3)+sin(pi/6+x)=0`
`<=> sin(x + π/6) = sin(2 x + (2 π)/3)`
`<=>` \(\left[ \begin{array}{l}
x + \frac{π}{6} = \frac{π}{3} - 2 x + 2 π n\\
x + \frac{π}{6} = \frac{2π}{3} + 2 x + 2 π n
\end{array} \right.\) (với `n\inZ`)
`<=>` \(\left[ \begin{array}{l}
3 x = \frac{π}{6}+2 π n
\\
-x = \frac{π}{2}+2 π n
\end{array} \right.\) (với `n\inZ`)
`<=>` \(\left[ \begin{array}{l}
x = \frac{π}{18}+\frac{2 π n}{3}
\\
x = \frac{-π}{2}-2 π n
\end{array} \right.\) (với `n\inZ`)
Vậy `S={π/18+(2 π n)/3; - π/2-2 π n | n\inZ}`
