Nghiệm của phương trình \(\begin{aligned} & \sin 2 x=(\sin x+\cos x-1)(2 \sin x+\cos x+2) \end{aligned}\) là:
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Lời giải:
Báo saiTXĐ:\(D=\mathbb{R}\)
\(\begin{aligned} & \sin 2 x=(\sin x+\cos x-1)(2 \sin x+\cos x+2) \\ \Leftrightarrow & \sin 2 x=\sin ^{2} x+3 \sin x \cos x+\cos x-1 \\ \Leftrightarrow & \sin ^{2} x-1+\sin x \cos x+\cos x=0 \\ \Leftrightarrow &(\sin x-1)(\sin x+1)+\cos x(\sin x+1)=0 \\ \Leftrightarrow &(\sin x+1)(\sin x+\cos x-1)=0 \\ \Leftrightarrow &\left[\begin{array}{l} \sin x=-1 \\ \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=1 \end{array}\right. \end{aligned}\)
\(\begin{aligned} &\Leftrightarrow\left[\begin{array}{l} \sin x=-1 \\ \cos \left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} \end{array}\right.\\ &\Leftrightarrow \quad\left[\begin{array}{l} x=-\frac{\pi}{2}+k_{1} 2 \pi \\ x-\frac{\pi}{4}=\frac{\pi}{4}+k_{2} 2 \pi \\ x-\frac{\pi}{4}=-\frac{\pi}{4}+k_{3} 2 \pi \end{array}\right.\\ &\Leftrightarrow \quad\left[\begin{array}{l} x=-\frac{\pi}{2}+k_{1} 2 \pi \\ x=\frac{\pi}{2}+k_{2} 2 \pi \\ x=k_{3} 2 \pi \end{array}\right. \end{aligned}\)
\(\Leftrightarrow \quad\left[\begin{array}{l} x=\frac{\pi}{2}+k \pi \\ x=k^{\prime} 2 \pi \end{array}\right.\)
