Question

\(cos2x+sin\left(x+\frac{\pi}{4}\right)=0\)

Answer 1

\(\cos2x=-\sin\left(x+\frac{\pi}{4}\right)=\sin\left(-x-\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{2}+x+\frac{\pi}{4}\right)=\cos\left(x+\frac{3}{4}\pi\right)\)

\(\Leftrightarrow\cos2x-\cos\left(x+\frac{3}{4}\pi\right)=0\)

\(\Leftrightarrow-2\sin\left(\frac{x}{2}-\frac{3}{8}\pi\right)\sin\left(\frac{3}{2}x+\frac{3}{8}\pi\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sin\left(\frac{x}{2}-\frac{3}{8}\pi\right)=0\\\sin\left(\frac{3}{2}x+\frac{3}{8}\pi\right)=0\end{matrix}\right.\Leftrightarrow....\)