$\begin{array}{l}g)\,\,\sin5x - 2\sin x\cos2x = 0\\ \Leftrightarrow \sin5x - (\sin3x - \sin x) = 0\\ \Leftrightarrow 2\sin3x\cos2x - \sin3x = 0\\ \Leftrightarrow \sin3x(2\cos x - 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin3x = 0\\\cos2x = \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}3x = k\pi\\2x = \pm \dfrac{\pi}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{3}\\x = \pm \dfrac{\pi}{7} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ i)\,\,\cos7x + 2\cos x\cos4x = 0\\ \Leftrightarrow \cos7x + (\cos5x + \cos3x) = 0\\ \Leftrightarrow 2\cos5x\cos2x + \cos5x = 0\\ \Leftrightarrow \cos5x(2\cos2x + 1) = 0\\ \Leftrightarrow \left[\begin{array}{l}\cos5x = 0\\\cos2x = - \dfrac{1}{2}\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}5x = \dfrac{\pi}{2} +k\pi\\2x = \pm \dfrac{2\pi}{3} + k2\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{10} +k\dfrac{\pi}{5}\\x = \pm \dfrac{\pi}{3} + k\pi\end{array}\right.\quad (k \in \Bbb Z) \end{array}$

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 !