Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = \dfrac{{{{\sin }^2}3x}}{{{{\sin }^2}x}} - \dfrac{{{{\cos }^2}3x}}{{{{\cos }^2}x}}\\
= \dfrac{{{{\sin }^2}3x.{{\cos }^2}x - {{\sin }^2}x.{{\cos }^2}3x}}{{{{\sin }^2}x.{{\cos }^2}x}}\\
= \dfrac{{{{\left( {\sin 3x.\cos x} \right)}^2} - {{\left( {\sin x.\cos 3x} \right)}^2}}}{{{{\sin }^2}x.{{\cos }^2}x}}\\
= \dfrac{{{{\left[ {\dfrac{1}{2}.\left( {\sin \left( {3x + x} \right) + \sin \left( {3x - x} \right)} \right)} \right]}^2} - {{\left[ {\dfrac{1}{2}\left( {\sin \left( {x + 3x} \right) + \sin \left( {x - 3x} \right)} \right)} \right]}^2}}}{{{{\sin }^2}x.{{\cos }^2}x}}\\
= \dfrac{{\dfrac{1}{4}.{{\left( {\sin 4x + \sin 2x} \right)}^2} - \dfrac{1}{4}.{{\left( {\sin 4x + \sin \left( { - 2x} \right)} \right)}^2}}}{{{{\sin }^2}x.{{\cos }^2}x}}\\
= \dfrac{{{{\left( {\sin 4x + \sin 2x} \right)}^2} - {{\left( {\sin 4x - \sin 2x} \right)}^2}}}{{4{{\sin }^2}x.{{\cos }^2}x}}\\
= \dfrac{{{{\sin }^2}4x + 2\sin 4x.\sin 2x + {{\sin }^2}2x - \left( {{{\sin }^2}4x - 2\sin 4x.\sin 2x + {{\sin }^2}2x} \right)}}{{{{\left( {2\sin x.\cos x} \right)}^2}}}\\
= \dfrac{{4\sin 4x.\sin 2x}}{{{{\sin }^2}2x}}\\
= \dfrac{{4.2.\sin 2x.\cos 2x.\sin 2x}}{{{{\sin }^2}2x}}\\
= 8\cos 2x
\end{array}\)
