\(\cos(x+\pi)=1+\sin(\dfrac{x}{2}+\dfrac{\pi}{2})\)
\(\Leftrightarrow \cos x\cos \pi-\sin x\sin \pi=1+\sin \dfrac{x}{2}\cos\dfrac{\pi}{2}+\cos\dfrac{x}{2}\sin\dfrac{\pi}{2}\)
\(\Leftrightarrow -\cos x=1+\cos\dfrac{x}{2}\)
\(\Leftrightarrow-2{\cos}^2 \dfrac{x}{2}+1=1+\cos\dfrac{x}{2}\)
\(\Leftrightarrow\cos\dfrac{x}{2}(1+2\cos\dfrac{x}{2})=0\)
\(\Leftrightarrow\cos\dfrac{x}{2}=0\) hoặc \(\cos\dfrac{x}{2}=-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi, k\in\mathbb{Z}\) hoặc \(\dfrac{x}{2}=\pm\dfrac{2\pi}{3}+k2\pi,k\in\mathbb Z\).
Vậy phương trình có nghiệm \(\dfrac{x}{2}=\dfrac{\pi}{2}+k\pi, k\in\mathbb{Z}\) và \(\dfrac{x}{2}=\pm\dfrac{2\pi}{3}+k2\pi,k\in\mathbb Z\).
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\dfrac{\pi}{2}\) hoặc \(x=\pm\dfrac{\pi}{3}+k\pi\) \(k\in\mathbb Z\).
