Giải phương trình: tan^2x.cot^22x.cot3x=tan^2x−cot^22x+cot3x

Hướng dẫn giải:

Điều kiện:  \( \left\{ \begin{align}  & \cos x\ne 0 \\  & \sin 2x\ne 0 \\  & \sin 3x\ne 0 \\ \end{align} \right. \) \( \Leftrightarrow \left\{ \begin{align} & \sin 2x\ne 0 \\  & \sin 3x\ne 0 \\ \end{align} \right. \).

Lúc đó: (*) \( \Leftrightarrow \cot 3x({{\tan }^{2}}x{{\cot }^{2}}2x-1)={{\tan }^{2}}x-{{\cot }^{2}}2x \)

 \( \Leftrightarrow \cot 3x\left[ \left( \frac{1-\cos 2x}{1+\cos 2x} \right)\left( \frac{1+\cos 4x}{1-\cos 4x} \right)-1 \right]=\frac{1-\cos 2x}{1+\cos 2x}-\frac{1+\cos 4x}{1-\cos 4x} \)

 \( \Leftrightarrow \cot 3x\left[ (1-\cos 2x)(1+\cos 4x)-(1+\cos 2x)(1-\cos 4x) \right] \)

 \(       =(1-\cos 2x)(1-\cos 4x)-(1+\cos 4x)(1+\cos 2x) \)

 \( \Leftrightarrow \cot 3x(2\cos 4x-2\cos 2x)=-2(\cos 4x+\cos 2x) \)

 \( \Leftrightarrow \frac{\cos 3x}{\sin 3x}(-4\sin 3x\sin x)=-4\cos 3x\cos x\Leftrightarrow \cos 3x\sin x=\cos 3x\cos x \) (do  \( \sin 3x\ne 0 \))

 \( \Leftrightarrow \left[ \begin{align}  & \cos 3x=0 \\  & \sin x=\cos x \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & 3x=\frac{\pi }{2}+k\pi  \\  & \tan x=1 \\ \end{align} \right. \) \( \Leftrightarrow \left[ \begin{align}  & x=\frac{\pi }{6}+\frac{k\pi }{3} \\  & x=\frac{\pi }{4}+h\pi  \\ \end{align} \right.,\text{ }(k,h\in \mathbb{Z}). \)

So sánh với điều kiện:  \( \sin 2x.\sin 3x\ne 0 \)

+ Khi  \( x=\frac{\pi }{6}+\frac{k\pi }{3} \) thì  \( \sin \left( \frac{\pi }{3}+\frac{k2\pi }{3} \right).\sin \left( \frac{\pi }{2}+k\pi  \right)\ne 0\Leftrightarrow \sin \left( \frac{1+2k}{3} \right)\pi \ne 0 \)

Luôn đúng  \( \forall k \) thỏa  \( 2k+1\ne 3m\text{ }(m\in \mathbb{Z}) \).

+ Khi  \( x=\frac{\pi }{4}+h\pi \)  thì  \( \sin \left( \frac{\pi }{2}+k2\pi  \right)\sin \left( \frac{3\pi }{4}+k3\pi  \right)=\pm \frac{\sqrt{2}}{2}\ne 0 \) luôn đúng.

Do đó: (*) \( \Leftrightarrow \left[ \begin{align}  & x=\frac{\pi }{6}+\frac{k\pi }{3},k\in \mathbb{Z}\wedge 2k\ne 3m-1\text{ }(m\in \mathbb{Z}) \\  & x=\frac{\pi }{4}+h\pi ,\text{ }h\in \mathbb{Z} \\ \end{align} \right. \).

Cách khác:

(*) \( \Leftrightarrow \cot 3x({{\tan }^{2}}x{{\cot }^{2}}2x-1)={{\tan }^{2}}x-{{\cot }^{2}}2x \)

 \( \Leftrightarrow \cot 3x=\frac{{{\tan }^{2}}x-{{\cot }^{2}}2x}{{{\tan }^{2}}x{{\cot }^{2}}2x-1}=\frac{{{\tan }^{2}}2x.{{\tan }^{2}}x-1}{{{\tan }^{2}}x-{{\tan }^{2}}2x} \)

 \( \Leftrightarrow \cot 3x=\frac{(1+\tan 2x.\tan x)(1-\tan 2x.\tan x)}{(\tan 2x-\tan x)(\tan 2x+\tan x)} \)

 \( \cot 3x=\cot x.\cot 3x\Leftrightarrow \cot 3x(\cot x-1)=0\Leftrightarrow \left[ \begin{align}  & \cot 3x=0 \\  & \cot x=1 \\ \end{align} \right. \).