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: \(\left\{\begin{array}{l} \cos x \neq 0 \\ \sin x \neq 0 \end{array} \Leftrightarrow \sin 2 x \neq 0 \Leftrightarrow x \neq \frac{k \pi}{2}, k \in \mathbb{Z}\right.\)
\(\begin{aligned} &2 \sin 3 x-\frac{1}{\sin x}=2 \cos 3 x+\frac{1}{\cos x} \Leftrightarrow 2(\sin 3 x-\cos 3 x)-\left(\frac{1}{\sin x}+\frac{1}{\cos x}\right)=0\\ &\Leftrightarrow 2\left(3 \sin x-4 \sin ^{3} x-4 \cos ^{3} x+3 \cos x\right)-\left(\frac{\cos x+\sin x}{\sin x \cos x}\right)=0\\ &\Leftrightarrow 6(\cos x+\sin x)-8(\cos x+\sin x)(1-\sin x \cos x)-\left(\frac{\cos x+\sin x}{\sin x \cos x}\right)=0\\ &\Leftrightarrow\left[\begin{array}{l} \cos x+\sin x=0 \,\,\,(1)\\ 6-8\left(1-\frac{1}{2} \sin 2 x\right)-\frac{2}{\sin 2 x}=0\,\,\,(2) \end{array}\right. \end{aligned}\)
+\(\text { (1) } \Leftrightarrow \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=0 \Leftrightarrow x=\frac{3 \pi}{4}+k \pi \Leftrightarrow x=-\frac{\pi}{4}+k \pi\)
+ \((2) \Leftrightarrow-2+4 \sin 2 x-\frac{2}{\sin 2 x}=0 \Leftrightarrow 2 \sin ^{2} 2 x-\sin 2 x-1=0\)
\(\Leftrightarrow\left[\begin{array}{l} \sin 2 x=1 \\ \sin 2 x=-\frac{1}{2} \end{array} \Leftrightarrow\left[\begin{array}{l} 2 x=\frac{\pi}{2}+k 2 \pi \\ 2 x=-\frac{\pi}{6}+k 2 \pi \\ 2 x=\frac{7 \pi}{6}+k 2 \pi \end{array} \Leftrightarrow\left[\begin{array}{l} x=\frac{\pi}{4}+k \pi \\ x=-\frac{\pi}{12}+k \pi \\ x=\frac{7 \pi}{12}+k \pi \end{array}\right.\right.\right.\)