Giải pt a) 2 sin^2 x = 1 + sin3x b) cot (x-bi/3) = tan2x

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1. 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.$cot(x-\dfrac{\pi}{3})=tan2x$

   $tan(\dfrac{\pi}{2}-(x-\dfrac{\pi}{3}))=tan2x$

   $tan(\dfrac{5\pi}{6}-x)=tan2x$

   $\rightarrow \dfrac{5\pi}{6}-x=2x+k\pi$

   $\rightarrow x=\dfrac{5\pi}{18}-\dfrac{k\pi}{3}$

   Trả lời