Sin3x -√3.cos3x=2cos5x

Đáp án:

$S=\{\ \dfrac{-5\pi}{12}+k\pi; -\dfrac{7\pi}{48}+k\dfrac{\pi}{4}|k\in Z \}$

Giải thích các bước giải:

 $sin3x-\sqrt{3}cos3x=2cos5x$

Chia cả hai vế cho 2 ta được:

$\dfrac{1}{2}sin3x-\dfrac{\sqrt{3}}{2}cos3x=cos5x$

$-(cos3x.cos\dfrac{\pi}{6}-sin3x.sin\dfrac{\pi}{6})=cos5x$

$-cos(3x+\dfrac{\pi}{6})=cos5x$

$cos(3x+\dfrac{\pi}{6})=-cos5x$

$cos(3x+\dfrac{\pi}{6})=cos(\pi+5x)$

\(\left[ \begin{array}{l}3x+\dfrac{\pi}{6}=\pi+5x+k2\pi\\3x+\dfrac{\pi}{6}=-\pi-5x+k2\pi\end{array} \right.\)

\(\left[ \begin{array}{l}-2x=\dfrac{5\pi}{6}+k2\pi\\8x=-\dfrac{7\pi}{6}+k2\pi\end{array} \right.\) 

\(\left[ \begin{array}{l}x=\dfrac{-5\pi}{12}+k\pi\\x=-\dfrac{7\pi}{48}+k\dfrac{\pi}{4}\end{array} \right.,k\in Z\) 

Vậy $S=\{\ \dfrac{-5\pi}{12}+k\pi; -\dfrac{7\pi}{48}+k\dfrac{\pi}{4}|k\in Z \}$