Phương trình \(2 \sin 3 x-\frac{1}{\sin x}=2 \cos 3 x+\frac{1}{\cos x}\) có nghiệm là:
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Lời giải:
Báo saiĐK: \(\sin 2 x \neq 0\)
Khi đó
\(\begin{array}{l} 2 \sin 3 x-\frac{1}{\sin x}=2 \cos 3 x+\frac{1}{\cos x} \Leftrightarrow 2(\sin 3 x-\cos 3 x)=\frac{1}{\cos x}+\frac{1}{\sin x} \\ \Leftrightarrow 2\left[\left(3 \sin x-4 \sin ^{3} x\right)-\left(4 \cos ^{3} x-3 \cos x\right)\right]=\frac{\sin x+\cos x}{\sin x \cos x} \\ \Leftrightarrow 2\left[3(\sin x+\cos x)-4\left(\sin ^{3} x+\cos ^{3} x\right)\right]=\frac{\sin x+\cos x}{\sin x \cos x} \\ \Leftrightarrow 2\left[3(\sin x+\cos x)-4(\sin x+\cos x)\left(\sin ^{2} x-\sin x \cos x+\cos ^{2} x\right)\right]=\frac{\sin x+\cos x}{\sin x \cos x} \\ \Leftrightarrow 2[3(\sin x+\cos x)-4(\sin x+\cos x)(1-\sin x \cos x)]=\frac{\sin x+\cos x}{\sin x \cos x} \end{array}\)
\(\begin{array}{l} \Leftrightarrow 2(\sin x+\cos x)[3-4(1-\sin x \cos x)]=\frac{\sin x+\cos x}{\sin x \cos x} \\ \Leftrightarrow(\sin x+\cos x)\left[6-8(1-\sin x \cos x)-\frac{1}{\sin x \cos x}\right]=0 \\ \Leftrightarrow(\sin x+\cos x)\left[-2+8 \sin x \cos x-\frac{1}{\sin x \cos x}\right]=0 \\ \Leftrightarrow \sqrt{2} \sin \left(x+\frac{\pi}{4}\right)\left[-2 \sin x \cos x+8(\sin x \cos x)^{2}-1\right]=0 \\ \Leftrightarrow \sin \left(x+\frac{\pi}{4}\right)\left[2 \sin ^{2} 2 x-\sin 2 x-1\right]=0 \end{array}\)
\(\Leftrightarrow\left[\begin{array}{l} \sin \left(x+\frac{\pi}{4}\right)=0 \\ \sin 2 x=1 \\ \sin 2 x=-\frac{1}{2} \end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{l} x+\frac{\pi}{4}=k \pi \\ 2 x=\frac{\pi}{2}+k 2 \pi \\ 2 x=-\frac{\pi}{6}+k 2 \pi \\ 2 x=\frac{7 \pi}{6}+k 2 \pi \end{array}\left(\begin{array}{l} k \in \mathbb{Z}) \\ \end{array}\right.\right.\)
\(\Leftrightarrow\left[\begin{array}{l} x=-\frac{\pi}{4}+k \pi \\ x=\frac{\pi}{4}+k \pi \\ x=-\frac{\pi}{12}+k \pi \\ x=\frac{7 \pi}{12}+k \pi \end{array}(k \in \mathbb{Z})\right.\)
