Giải thích các bước giải:
a.$2sin^2x=1+sin3x$
$\rightarrow 2sin^2x-1=sin3x$
$\rightarrow -cos2x=cos(\dfrac{\pi}{2}-3x)$
$\rightarrow cos(\pi-2x)=cos(3x-\dfrac{\pi}{2})$
$\rightarrow \pi-2x=3x-\dfrac{\pi}{2}+k2\pi$ hoặc $\pi-2x=-3x+\dfrac{\pi}{2}+k2\pi$
$\rightarrow x=\dfrac{3\pi}{10}-\dfrac{k2\pi}{5}$ hoặc $x=\dfrac{-\pi}{2}+k2\pi$
b.$cot(x-\dfrac{\pi}{3})=tan2x$
$tan(\dfrac{\pi}{2}-(x-\dfrac{\pi}{3}))=tan2x$
$tan(\dfrac{5\pi}{6}-x)=tan2x$
$\rightarrow \dfrac{5\pi}{6}-x=2x+k\pi$
$\rightarrow x=\dfrac{5\pi}{18}-\dfrac{k\pi}{3}$
