g) $sin5x-2sinx.cos2x=0$
<=>$sin5x-[sin3x+sin(-x)]=0$
<=>$sin5x-sin3x+sinx=0$
<=>$2sin3x.cos2x-sin3x=0$
<=>$sin3x.(2cos2x-1)=0$
<=>$\left[ \begin{array}{l}sin3x=0\\2cos2x-1=0\end{array} \right.$
<=>$\left[ \begin{array}{l}3x=k\pi\\cos2x=\dfrac{1}{2}=cos\dfrac{\pi}{3}\end{array} \right.$
<=>$\left[ \begin{array}{l}3x=k\pi\\\left[ \begin{array}{l}2x=\dfrac{\pi}{3}+k2\pi\\2x=-\dfrac{\pi}{3}+k2\pi\end{array} \right.\end{array} \right.$
<=>$\left[ \begin{array}{l}3x=k\pi\\\left[ \begin{array}{l}x=\dfrac{\pi}{6}+k\pi\\x=-\dfrac{\pi}{6}+k\pi\end{array} \right.\end{array} \right.(k∈\mathbb{Z})$
i) $cos7x+2cosx.cos4x=0$
<=>$cos7x+cos5x+cos(-3x)=0$
<=>$cos7x+cos3x+cos5x=0$
<=>$2cos5x.cos2x+cos5x=0$
<=>$cos5x.(2cos2x+1)=0$
bạn giải như trên nha :(( dài quá sắp đi hc r !
