cos6x + sin3x-1=0
<=>$1-2sin^{3}x+sin3x-1=0$
<=>$sin3x(-2sin3x+1)=0$
<=>$\left[ \begin{array}{l}sin3x=0\\-2sin3x+1=0\end{array} \right.$
<=> $\left[ \begin{array}{l}3x=kπ\\sin3x=\frac{1}{2}\end{array} \right.$
<=> $\left[ \begin{array}{l}x=\frac{kπ}{3}\\sin3x=sin\frac{π}{6}\end{array} \right.$
<=>$\left[ \begin{array}{l}x=\frac{kπ}{3}\\\left[ \begin{array}{l}3x=\frac{π}{6}+k2π\\3x=π-\frac{π}{6}+k2π\end{array} \right. \end{array} \right.$
<=> $\left[ \begin{array}{l}x=\frac{kπ}{3}\\\left[ \begin{array}{l}x=\frac{π}{16}+\frac{k2π}{3}\\x=\frac{5π}{18}+\frac{k2π}{3}\end{array} \right. \end{array} \right.(K∈Z)$
Chúc bạn học tốt!
