Đáp án:
$\begin{array}{l}
\frac{1}{{4{{\cos }^2}x}} = \frac{4}{{{{\sin }^2}2x}} - \frac{1}{{4{{\sin }^2}x}}\\
\Leftrightarrow \frac{1}{{4{{\cos }^2}x}} + \frac{1}{{4{{\sin }^2}x}} = \frac{4}{{{{\left( {2\sin x.\cos x} \right)}^2}}}\\
\Leftrightarrow \frac{{{{\sin }^2}x + {{\cos }^2}x}}{{4{{\sin }^2}x.{{\cos }^2}x}} = \frac{4}{{4.{{\sin }^2}x.{{\cos }^2}x}}\\
\Leftrightarrow \frac{1}{{4{{\sin }^2}x.{{\cos }^2}x}} = \frac{4}{{4{{\sin }^2}x.{{\cos }^2}x}}\left( {ktm} \right)
\end{array}$
=> Em xem lại đề xem đã chép đúng chưa nhé.
