ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(sin\left(x+\dfrac{\pi}{4}\right)\left(1+sinx+cos2x\right)=\dfrac{1}{\sqrt{2}}cosx\left(1+\dfrac{sinx}{cosx}\right)\)
\(\Rightarrow sin\left(x+\dfrac{\pi}{4}\right)\left(1+sinx+cos2x\right)=\dfrac{1}{\sqrt{2}}\left(sinx+cosx\right)\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)\left(1+sinx+cos2x\right)=sin\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\Rightarrow x=-\dfrac{\pi}{4}+k\pi\\1+sinx+cos2x=1\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow sinx+cos2x=0\)
\(\Leftrightarrow-2sin^2x+sinx+1=0\Rightarrow\left[{}\begin{matrix}sinx=1\\sinx=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=-\dfrac{\pi}{6}+k2\pi\\x=\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
