Giải phương trình sau: \(\sin 5x+\cos 5x=-1\)

Ta có \(\sin 5x+\cos 5x=-1\)

\(\Leftrightarrow \dfrac{1}{\sqrt{2}}\cos 5x+\dfrac{1}{\sqrt{2}}\sin 5x=-\dfrac{1}{\sqrt{2}}\)

Trong đó \(\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}\), \(\sin \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}\) và \(\sin {\left({-\dfrac{\pi}{4}}\right)}=-\dfrac{1}{\sqrt{2}}\)

Ta thu được phương trình

\(\cos \dfrac{\pi}{4}\sin 5x+\sin\dfrac{\pi}{4}\cos 5x=\sin {\left({-\dfrac{\pi}{4}}\right)}\)

\(\Leftrightarrow \sin (5x+\dfrac{\pi}{4})=\sin {\left({-\dfrac{\pi}{4}}\right)}\Leftrightarrow\)

\(\left[ \begin{array}{l} 5x+\dfrac{\pi}{4} = -\dfrac{\pi}{4}+k2\pi ,k \in \mathbb{Z}\\5x+\dfrac{\pi}{4}=\pi-{\left({-\dfrac{\pi}{4}}\right)}+k2\pi ,k \in \mathbb{Z}\end{array} \right. \)

\(\Leftrightarrow\left[ \begin{array}{l} x= -\dfrac{\pi}{10}+k\dfrac{2\pi}{5} ,k \in \mathbb{Z}\\x=\dfrac{\pi}{5}+k\dfrac{2\pi}{5} ,k \in \mathbb{Z}\end{array} \right. \)

Vậy phương trình có nghiệm là \(x= -\dfrac{\pi}{10}+k\dfrac{2\pi}{5} ,k \in \mathbb{Z}\) và \(x=\dfrac{\pi}{5}+k\dfrac{2\pi}{5} ,k \in \mathbb{Z}\).