Đề sai nhiều chỗ vậy, lần sau ghi đúng đề đi.
\(cos3x+sin7x=2sin^2\left(\dfrac{\pi}{4}-\dfrac{5x}{2}\right)+2cos^2\dfrac{9x}{2}\)
\(\Leftrightarrow cos3x+sin7x=cos\left(\dfrac{\pi}{2}-5x\right)+1-2cos^2\dfrac{9x}{2}\)
\(\Leftrightarrow cos3x+sin7x=sin5x-cos9x\)
\(\Leftrightarrow2cos6x.cos3x+2cos6x.sinx=0\)
\(\Leftrightarrow2cos6x.\left(cos3x+sinx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos3x+sinx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos3x+cos\left(\dfrac{\pi}{2}-x\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\2cos\left(\dfrac{\pi}{4}+x\right).cos\left(2x-\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos6x=0\\cos\left(\dfrac{\pi}{4}+x\right)=0\\cos\left(2x-\dfrac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=\dfrac{\pi}{2}+k\pi\\\dfrac{\pi}{4}+x=\dfrac{\pi}{2}+k\pi\\2x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+\dfrac{k\pi}{6}\\x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{3\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
