Giải phương trình: sin^42x+cos^42x/tan(π/4−x)tan(π/4+π)=cos^44x

Hướng dẫn giải:

Điều kiện: \(\left\{ \begin{align}  & \sin \left( \frac{\pi }{4}-x \right)\cos x\left( \frac{\pi }{4}-x \right)\ne 0 \\  & \sin \left( \frac{\pi }{4}+x \right)\cos x\left( \frac{\pi }{4}+x \right)\ne 0 \\ \end{align} \right.\)\(\Leftrightarrow \left\{ \begin{align}  & \sin \left( \frac{\pi }{2}-2x \right)\ne 0 \\  & \sin \left( \frac{\pi }{2}+2x \right)\ne 0 \\ \end{align} \right.\Leftrightarrow \cos 2x\ne 0\Leftrightarrow \sin 2x\ne \pm 1\).

Do:  \( \tan \left( \frac{\pi }{4}-x \right)\tan \left( \frac{\pi }{4}+x \right)=\frac{1-\tan x}{1+\tan x}.\frac{1+\tan x}{1-\tan x}=1 \).

Khi  \( \cos 2x\ne 0 \) thì:

(*) \( \Leftrightarrow {{\sin }^{4}}2x+{{\cos }^{4}}2x={{\cos }^{4}}4x\Leftrightarrow 1-2{{\sin }^{2}}2xco{{s}^{2}}2x={{\cos }^{4}}4x \)

 \( \Leftrightarrow 1-\frac{1}{2}si{{n}^{2}}4x={{\cos }^{4}}4x\Leftrightarrow 1-\frac{1}{2}(1-{{\cos }^{2}}4x)={{\cos }^{4}}4x \)

 \( \Leftrightarrow 2{{\cos }^{4}}4x-{{\cos }^{2}}4x-1=0\Leftrightarrow \left[ \begin{align}  & {{\cos }^{2}}4x=1\text{ }(n) \\  & {{\cos }^{2}}4x=-\frac{1}{2}\text{ }(\ell ) \\ \end{align} \right.\Leftrightarrow 1-{{\sin }^{2}}4x=1 \)

 \( \Leftrightarrow \sin 4x=0\Leftrightarrow 2\sin 2x\cos 2x=0\Leftrightarrow \sin 2x=0\text{ }(do\text{ }\cos 2x\ne 0) \)

 \( \Leftrightarrow 2x=k\pi \Leftrightarrow x=\frac{k\pi }{2},\text{ }k\in \mathbb{Z} \).