Chọn phương án A.

a) $\sin3x+\cos3x=\sqrt{2}\cos2x$
Chia hai vế cho $\sqrt{2}$ ta được $$\begin{array}{lll}
&\dfrac{1}{\sqrt{2}}\sin3x+\dfrac{1}{\sqrt{2}}\cos3x&=\cos2x\\
\Leftrightarrow&\sin\dfrac{\pi}{4}\sin3x+\cos\dfrac{\pi}{4}\cos3x&=\cos2x\\
\Leftrightarrow&\cos\left(3x-\dfrac{\pi}{4}\right)&=\cos2x\\
\Leftrightarrow&\left[\begin{array}{l}3x-\dfrac{\pi}{4}=2x+k2\pi\\ 3x-\dfrac{\pi}{4}=-2x+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\\ 5x=\dfrac{\pi}{4}+k2\pi\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\dfrac{\pi}{4}+k2\pi\\ x=\dfrac{\pi}{20}+k\dfrac{2\pi}{5}\end{array}\right.\,(k\in\mathbb{Z})
\end{array}$$

b) $(2\sin x-\cos x)(1+\cos x)=\sin^2x$
Ta có $\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x)$.

Do đó, phương trình đã cho trở thành $$\begin{array}{lll}
&(2\sin x-\cos x)(1+\cos x)&=(1-\cos x)(1+\cos x)\\
\Leftrightarrow&(1+\cos x)\big[(2\sin x-\cos x)-(1-\cos x)\big]&=0\\
\Leftrightarrow&(1+\cos x)\big[2\sin x-1\big]&=0\\
\Leftrightarrow&\left[\begin{array}{l}1+\cos x=0\\ 2\sin x-1=0\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}\cos x=-1\\ \sin x=\dfrac{1}{2}\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{l}x=\pi+k2\pi\\ x=\dfrac{\pi}{6}+k2\pi\\ x=\dfrac{5\pi}{6}+k2\pi\end{array}\right.\,(k\in\mathbb{Z})
\end{array}$$