Sửa đề: cosx+sinx=1
=>\(\sqrt{2}\cdot sin\left(x+\dfrac{pi}{4}\right)=1\)
=>\(sin\left(x+\dfrac{pi}{4}\right)=\dfrac{1}{\sqrt{2}}\)
=>\(\left[{}\begin{matrix}x+\dfrac{pi}{4}=\dfrac{pi}{4}+k2pi\\x+\dfrac{pi}{4}=\dfrac{3}{4}pi+k2pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=k2pi\\x=pi+k2pi\end{matrix}\right.\)