ta có : \(sin\left(2x+\dfrac{3\pi}{4}\right)+cosx=0\Leftrightarrow cosx=sin\left(-2x-\dfrac{3\pi}{4}\right)\)
\(\Leftrightarrow cosx=sin\left(\dfrac{\pi}{2}-\left(2x+\dfrac{5\pi}{4}\right)\right)\Leftrightarrow cosx=cos\left(2x+\dfrac{5\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2x+\dfrac{5\pi}{4}+k2\pi\\x=-2x-\dfrac{5\pi}{4}+k2\pi\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{5\pi}{4}+k2\pi\\3x=\dfrac{-5\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5\pi}{4}-k2\pi\\x=\dfrac{-5\pi}{12}+\dfrac{2}{3}k\pi\end{matrix}\right.\) \(\left(k\in Z\right)\)
vậy phương trình có 2 hệ nghiệm : \(x=\dfrac{-5\pi}{4}-k2\pi\) và \(x=\dfrac{-5\pi}{12}+\dfrac{2}{3}k\pi\)
