cos3x=sin(\(\dfrac{\pi}{2}\)-3x)

\(\Leftrightarrow\)sin(\(\dfrac{\pi}{2}\)-3x)=sin2x

\(\Leftrightarrow\)2x=\(\dfrac{\pi}{2}\)-3x+k2\(\pi\) or 2x=3x-\(\dfrac{\pi}{2}\)+k2\(\pi\)

\(\begin{array}{l}a)\;sin2x + cos3x = 0\\ \Leftrightarrow cos\left( {\frac{\pi }{2} - 2x} \right) + cos3x = 0\\ \Leftrightarrow cos\left( {\frac{\pi }{2} - 2x} \right) =  - cos3x\\ \Leftrightarrow cos\left( {\frac{\pi }{2} - 2x} \right) = cos\left( {\pi  - 3x} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{\pi }{2} - 2x = \pi  - 3x + k2\pi \\\frac{\pi }{2} - 2x =  - \pi  + 3x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{2} + k2\pi \\x = \frac{{3\pi }}{{10}} + k\frac{{2\pi }}{5}\end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}b)\;sinx.cosx = \frac{{\sqrt 2 }}{4}\\ \Leftrightarrow \frac{1}{2}\;sin2x = \frac{{\sqrt 2 }}{4}\\ \Leftrightarrow sin2x = \frac{{\sqrt 2 }}{2} = sin\left( {\frac{\pi }{4}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = \frac{\pi }{4} + k2\pi \\2x = \pi  - \frac{\pi }{4} + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{8} + k\pi \\x = \frac{{3\pi }}{8} + k\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(\begin{array}{l}c)\;sinx + sin2x = 0\\ \Leftrightarrow sinx =  - sin2x\\ \Leftrightarrow sinx = sin( - 2x)\\ \Leftrightarrow \left[ \begin{array}{l}x =  - 2x + k2\pi \\x = \pi  + 2x + k2\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = k\frac{{2\pi }}{3}\\x =  - \pi  + k2\pi \end{array} \right.\left( {k \in \mathbb{Z}} \right)\end{array}\)

\(A=cos2x+sin4x-cos6x\)

\(=\left(cos2x-cos6x\right)+sin4x=-2.sin4x.sin\left(-2x\right)+sin4x\)

\(=2sin4x.sin2x+sin4x=sin4x\left(2sin2x+1\right)\)

\(B=sinx-sin2x+sin5x+sin8x\)

\(=\left(sin5x+sinx\right)+\left(sin8x-sin2x\right)\)

\(=2.sin3x.cos2x+2.sin3x.cos5x\)

\(=2sin3x\left(cos2x+cos5x\right)\)