$cos^2(x+\dfrac{\pi}{4})+sin(x+\dfrac{\pi}{4})+1=0$
$↔ 1-sin^2(x+\dfrac{\pi}{4})+sin(x+\dfrac{\pi}{4})+1=0$
$↔ sin^2(x+\dfrac{\pi}{4})-sin(x+\dfrac{\pi}{4})-2=0$
$↔ \left[ \begin{array}{l}sin(x+\dfrac{\pi}{4})=-1\\sin(x+\dfrac{\pi}{4})=2\end{array} \right.$
Loại $sin(x+\dfrac{\pi}{4})=2$ vì $sin(x+\dfrac{\pi}{4})∈[-1;1]$
$↔ sin(x+\dfrac{\pi}{4})=-1$
$↔ x+\dfrac{\pi}{4}=\dfrac{-\pi}{2}+k2\pi$
$↔ x=\dfrac{-3\pi}{4}+k2\pi$.
