Question

Answer 1

\(\Leftrightarrow\dfrac{\sqrt{2}}{2}\sin2x-\dfrac{\sqrt{2}}{2}\cos2x=\cos x\)

\(\Leftrightarrow\sin\left(2x-\dfrac{\pi}{4}\right)=\sin\left(\dfrac{\pi}{2}-x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{4}=\dfrac{\pi}{2}-x+k2\pi\\2x-\dfrac{\pi}{4}=\pi-\dfrac{\pi}{2}+x+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\\x=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)

\(\Rightarrow2\alpha-\beta=\dfrac{2.3\pi}{4}-\dfrac{\pi}{4}=\dfrac{5\pi}{4}\)