Giải phương trình \(\left(1+\sqrt{2}\right)\left(\sin x+\cos x\right)-\sin2x-1-\sqrt{2}=0\).
\(\left[\begin{array}{nghiempt}x=2k\pi\\x=\frac{\pi}{2}+2k\pi\\x=\frac{\pi}{4}+2k\pi\end{array}\right.\) với \(k\in Z.\)\(\left[\begin{array}{nghiempt}x=2k\pi\\x=\frac{\pi}{2}+2k\pi\end{array}\right.\) với \(k\in Z.\)\(x=\pm\frac{\pi}{4}+k\pi\) với \(k\in Z.\)\(x=\frac{\pi}{2}+k\pi\) với \(k\in Z.\)Hướng dẫn giải:Phương trình tương đương với:
\(\left(1+\sqrt{2}\right)\left(\sin x+\cos x\right)-2\sin x\cos x-1-\sqrt{2}=0\)
Đặt \(t=\sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)\), điều kiện: \(-\sqrt{2}\le t\le\sqrt{2}\).
Khi đó: \(t^2=\left(\sin x+\cos x\right)^2=\sin^2x+2\sin x\cos x+\cos^2x=1+2\sin x\cos x\)
Suy ra \(2\sin x\cos x=t^2-1\)
Thay vào phương trình ta có:
\(\left(1+\sqrt{2}\right)t-\left(t^2-1\right)-1-\sqrt{2}=0\)
\(t^2-\left(1+\sqrt{2}\right)t+\sqrt{2}=0\)
\(\left[\begin{array}{nghiempt}t=1\\t=\sqrt{2}\end{array}\right.\)
Suy ra: \(\left[\begin{array}{nghiempt}\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=1\\\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sin\left(x+\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\\\sin\left(x+\frac{\pi}{4}\right)=1\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sin\left(x+\frac{\pi}{4}\right)=\sin\frac{\pi}{4}\\x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi;k\in\mathbb{Z}\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+\frac{\pi}{4}=\frac{\pi}{4}+2k\pi\\x+\frac{\pi}{4}=\pi-\frac{\pi}{4}+2k\pi\\x+\frac{\pi}{4}=\frac{\pi}{2}+2k\pi\end{array}\right.\) (với \(k\in\mathbb{Z}\))
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2k\pi\\x=\frac{\pi}{2}+2k\pi\\x=\frac{\pi}{4}+2k\pi\end{array}\right.\) với \(k\in Z.\)