sin5x + sin3x - cosx =0

Đáp án:

Giải thích các bước giải:

 $sin5x+sin3x-cosx=0$

$2sin4x.cosx-cosx=0$

$cosx(2sin4x-1)=0$

\(\left[ \begin{array}{l}cosx=0\\2sin4x-1=0\end{array} \right.\) 

\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\\sin4x=\dfrac{1}{2}\end{array} \right.\) 

\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\\\left[ \begin{array}{l}4x=\dfrac{\pi}{6}+k2\pi\\4x=\pi-\dfrac{\pi}{6}+k2\pi\end{array} \right. \end{array} \right.\)

\(\left[ \begin{array}{l}x=\dfrac{\pi}{2}+k\pi\\\left[ \begin{array}{l}x=\dfrac{\pi}{24}+\dfrac{k\pi}{2}\\x=\dfrac{5\pi}{24}+\dfrac{k\pi}{2}\end{array} \right. \end{array} \right. , k\in Z\)

Vậy tập nghệm của phương trình là :

$S=\{\  \dfrac{\pi}{2}+k\pi;\dfrac{\pi}{24}+\dfrac{k\pi}{2};\dfrac{5\pi}{24}+\dfrac{k\pi}{2} |k\in Z \}$