Giải phương trình: sin^4x+cos^4x=7/8cot(x+π/3)cot(π/6−x)

Hướng dẫn giải:

Điều kiện:  \( \left\{ \begin{align} & \sin \left( x+\frac{\pi }{3} \right)\ne 0 \\  & \sin \left( \frac{\pi }{6}-x \right)\ne 0 \\ \end{align} \right. \) \( \Leftrightarrow \left\{ \begin{align} & \sin \left( x+\frac{\pi }{3} \right)\ne 0 \\  & \cos \left( x+\frac{\pi }{3} \right)\ne 0 \\ \end{align} \right.\Leftrightarrow \sin \left( 2x+\frac{2\pi }{3} \right)\ne 0 \)

 \( \Leftrightarrow -\frac{1}{2}\sin 2x+\frac{\sqrt{3}}{2}\cos 2x\ne 0\Leftrightarrow \tan 2x\ne \sqrt{3} \).

Ta có:  \( {{\sin }^{4}}x+{{\cos }^{4}}x={{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x.co{{s}^{2}}x=1-\frac{1}{2}{{\sin }^{2}}2x \).

Và  \( \cot \left( x+\frac{\pi }{3} \right).\cot \left( \frac{\pi }{6}-x \right)=\cot \left( x+\frac{\pi }{3} \right).\tan \left( \frac{\pi }{3}+x \right)=1 \).

Lúc đó: (*) \( \Leftrightarrow 1-\frac{1}{2}{{\sin }^{2}}2x=\frac{7}{8}\Leftrightarrow -\frac{1}{4}(1-\cos 4x)=-\frac{1}{8}\Leftrightarrow \cos 4x=\frac{1}{2} \)

 \( \Leftrightarrow 4x=\pm \frac{\pi }{3}+k2\pi \Leftrightarrow x=\pm \frac{\pi }{12}+\frac{k\pi }{2} \) (nhận do  \( \tan 2x=\pm \frac{\sqrt{3}}{3}\ne \sqrt{3} \)).