Giải phương trình: 48−1/cos^4x−2/sin^2x(1+cot2xcotx)=0

Hướng dẫn giải:

Điều kiện:  \( \sin 2x\ne 0 \).

Ta có:  \( 1+\cot 2x\cot x=1+\frac{\cos 2x}{\sin 2x}.\frac{\cos x}{\sin x}=\frac{\sin 2x\sin x+\cos 2x\cos x}{\sin x\sin 2x} \)

 \( =\frac{\cos x}{2{{\sin }^{2}}x\cos x}=\frac{1}{2{{\sin }^{2}}x}\text{ }(do\text{ }\cos x\ne 0) \).

Lúc đó (*) \( \Leftrightarrow 48-\frac{1}{{{\cos }^{4}}x}-\frac{1}{{{\sin }^{4}}x}=0\Leftrightarrow 48=\frac{1}{{{\cos }^{4}}x}+\frac{1}{{{\sin }^{4}}x}=\frac{{{\sin }^{4}}x+{{\cos }^{4}}x}{{{\sin }^{4}}x{{\cos }^{4}}x} \)

 \( \Leftrightarrow 48{{\sin }^{4}}xco{{s}^{4}}x={{\sin }^{4}}x+co{{s}^{4}}x\Leftrightarrow 3{{\sin }^{4}}2x=1-2{{\sin }^{2}}xco{{s}^{2}}x \)

 \( \Leftrightarrow 3{{\sin }^{4}}2x+\frac{1}{2}{{\sin }^{2}}2x-1=0\Leftrightarrow \left[ \begin{align}  & {{\sin }^{2}}2x=-\frac{2}{3}\text{ }(\ell ) \\ & {{\sin }^{2}}2x=\frac{1}{2}\text{ }(n) \\ \end{align} \right. \)

 \( \Leftrightarrow \frac{1}{2}(1-\cos 4x)=\frac{1}{2}\Leftrightarrow \cos 4x=0\Leftrightarrow 4x=\frac{\pi }{2}+k\pi \Leftrightarrow x=\frac{\pi }{8}+\frac{k\pi }{4},\text{ }k\in \mathbb{Z} \).