b.$$sin(\dfrac{\pi}{2}+3x)-sin(\pi-5x)=\sqrt{3}(cos5x-sin3x)$$
$$\leftrightarrow sin(\dfrac{\pi}{2}-(-3x))-sin(\pi-5x)=\sqrt{3}(cos5x-sin3x)$$
$$\leftrightarrow cos3x -sin5x=\sqrt{3}(cos5x-sin3x)$$
$$\leftrightarrow \dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cos5x+\dfrac{1}{2}sin5x$$
$$\leftrightarrow sin(3x+\dfrac{\pi }{6})=sin(5x+\dfrac{\pi }{3})$$
$$\leftrightarrow \begin{bmatrix}& 3x+\dfrac{\pi }{6} =5x+\dfrac{\pi }{3}+k2\pi & \\ & 3x+\dfrac{\pi }{6} = \pi -5x-\dfrac{\pi }{3}+k2\pi & \end{bmatrix}$$
$$ \leftrightarrow \begin{bmatrix}& x= \dfrac{-\pi }{12}-k\pi & \\ & x=\dfrac{\pi }{16}+\dfrac{1}{4} k\pi & \end{bmatrix}$$