Chắc là bằng 0????
\(sin\left(\frac{\pi}{2}cos2x\right)=0\)
\(\Leftrightarrow\frac{\pi}{2}cos2x=k\pi\)
\(\Leftrightarrow cos2x=2k\)
Do \(-1\le cos2x\le1\Rightarrow-1\le2k\le1\Rightarrow k=0\)
\(\Rightarrow cos2x=0\Rightarrow x=\frac{\pi}{4}+\frac{k\pi}{2}\)