Đáp án: `x=(kπ)/2 ; x = π/3 +kπ (k \in \mathbbZ)`
Giải thích các bước giải:
$sinx+sin2x+sin3x=0$
$⇔ (sinx+sin3x)+sin2x=0$
$⇔-2sin2xcosx +sin2x=0$
$⇔sin2x(-2cosx+1)=0$
$⇔$ \(\left[ \begin{array}{l}sin2x=0\\cosx=\dfrac{1}{2}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=kπ\\x=\pm \dfrac{π}{3}+k2π\end{array} \right.\)
`<=> ` \(\left[ \begin{array}{l}x=\dfrac{kπ}{2}\\x=\pm \dfrac{π}{3}+k2π\end{array} \right.\)
Vậy `x=(kπ)/2 ; x = π/3 +kπ (k \in \mathbbZ)`
