$cos(x-120^o)+sin(2x+400^o)=0$
$⇔cos[90-(x-30^o)]=-sin(2x+400^o)$
$⇔sin(x-30^o)=sin(-2x-400^o)$
$⇔$\(\left[ \begin{array}{l}x-30^o=-2x-400^o+k.360^o\\x-30^o=180^o+2x+400^o+k.360^o\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}3x=-370^o+k.360^o\\-x=610^o+k.360^o\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=-\dfrac{370^o}{3}+k.120^o\\x=-610^o-k.360^o\end{array} \right.\)
