$f)cos2xcosx+sinxcos3x=sin2xsinx-sin3xcosx$

$⇔-sin2xsinx+cos2xcosx=-(sinxcos3x+sin3xcosx)$

$⇔cos(2x+x)=-sin(3x+x)$

$⇔cos3x=-sin4x$

$⇔cos3x=cos(\dfrac{\pi}{2}+4x)$

$⇔$\(\left[ \begin{array}{l}3x=\dfrac{\pi}{2}+4x+k2\pi\\3x=-\dfrac{\pi}{2}-4x+k2\pi\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=-\dfrac{\pi}{2}-k2\pi\\x=-\dfrac{\pi}{14}+\dfrac{k2\pi}{7}\end{array} \right.\) $(k∈Z)$

$g)sinx+sin2x+sin3x=1+cosx+cos2x$

$⇔sin2x+(sin3x+sinx)=1+cosx+2cos^2x-1$

$⇔sin2x+2sin2xcosx=1+cosx+2cos^2x-1$

$⇔sin2x(1+2cosx)-cosx(1+2cosx)=0$

$⇔(1+2cosx)(sin2x-cosx)=0$

$1+2cosx=0$

$⇔cosx=-\dfrac{1}{2}=cos(\dfrac{2\pi}{3})$

$⇔x=±\dfrac{2\pi}{3}+k2\pi(k∈Z)$

$sin2x-cosx=0$

$⇔sin2x=cosx$

$⇔sin2x=sin(\dfrac{\pi}{2}-x)$

$⇔$\(\left[ \begin{array}{l}2x=\dfrac{\pi}{2}-x+k2\pi\\2x=\pi-\dfrac{\pi}{2}+x+k2\pi\end{array} \right.\) 

$⇔$\(\left[ \begin{array}{l}x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\\x=\dfrac{\pi}{2}+k2\pi\end{array} \right.\) $(k∈Z)$

$\begin{array}{l}
f)\cos 2x\cos x + \sin x\cos 3x = \sin 2x\sin x - \sin 3x\cos x\\
 \Leftrightarrow \cos 2x\cos x - \sin 2x\sin x =  - \left( {\sin x\cos 3x + \sin 3x\cos x} \right)\\
 \Leftrightarrow \cos \left( {2x + x} \right) =  - \sin \left( {3x + x} \right)\\
 \Leftrightarrow \cos \left( {2x + x} \right) =  - \sin \left( {4x} \right)\\
 \Rightarrow \cos 3x = \sin \left( { - 4x} \right)\\
 \Leftrightarrow \cos 3x = \cos \left( {\dfrac{\pi }{2} + 4x} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
4x + \dfrac{\pi }{2} = 3x + k2\pi \\
4x + \dfrac{\pi }{2} =  - 3x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{2} + k2\pi \\
x = \dfrac{\pi }{{14}} + \dfrac{{k2\pi }}{7}
\end{array} \right.\left( {k \in Z} \right)\\
g)\sin x + \sin 2x + \sin 3x = 1 + \cos x + \cos 2x\\
 \Leftrightarrow \left( {\sin x + \sin 3x} \right) + \sin 2x = 1 + \cos x + 2{\cos ^2}x - 1\\
 \Leftrightarrow 2\sin 2x\cos x + \sin 2x = \cos x\left( {1 + 2\cos x} \right)\\
 \Leftrightarrow \sin 2x\left( {2\cos x + 1} \right) = \cos x\left( {1 + \cos 2x} \right)\\
 \Leftrightarrow \left( {1 + 2\cos 2x} \right)\left( {\sin 2x - \cos x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos 2x =  - \dfrac{1}{2}\\
\sin 2x = \cos x
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x =  \pm \dfrac{{2\pi }}{3} + k2\pi \\
\dfrac{\pi }{2} - 2x = x + k2\pi \\
\dfrac{\pi }{2} - 2x =  - x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  \pm \dfrac{{2\pi }}{3} + k2\pi \\
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\\
x = \dfrac{\pi }{2} - k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ \pm 2\pi }}{3} + k2\pi \\
x = \dfrac{\pi }{2} + k\pi \\
x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}
\end{array} \right.\left( {k \in Z} \right)\\
h){\sin ^2}x + {\cos ^2}2x = {\sin ^2}2x + {\cos ^2}x\\
 \Leftrightarrow {\cos ^2}x - {\sin ^2}x = {\cos ^2}2x - {\sin ^2}2x\\
 \Leftrightarrow \cos 2x = \cos 4x\\
 \Leftrightarrow \left[ \begin{array}{l}
4x = 2x + k2\pi \\
4x =  - 2x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = k\pi \\
x = \dfrac{{k\pi }}{3}
\end{array} \right. \Rightarrow x = \dfrac{{k\pi }}{3}\left( {k \in \mathbb{Z}} \right)
\end{array}$