Đáp án:
\(\dfrac{{1 - 8{{\cos }^2}x}}{{2 - 8{{\cos }^2}x}}\)
Giải thích các bước giải:
\(\begin{array}{l}
\cot x = \dfrac{1}{2}\\
\to \dfrac{{\cos x}}{{\sin x}} = \dfrac{1}{2}\\
\to \sin x = 2\cos x\\
Thay:\sin x = 2\cos x\\
{G_3} = \dfrac{{8{{\cos }^3}x - 2{{\left( {2\cos x} \right)}^3} + \cos x}}{{2\cos x - {{\left( {2\cos x} \right)}^3}}}\\
= \dfrac{{8{{\cos }^3}x - 16{{\cos }^3}x + \cos x}}{{2\cos x - 8{{\cos }^3}x}}\\
= \dfrac{{ - 8{{\cos }^3}x + \cos x}}{{2\cos x - 8{{\cos }^3}x}}\\
= \dfrac{{\cos x\left( {1 - 8{{\cos }^2}x} \right)}}{{\cos x\left( {2 - 8{{\cos }^2}x} \right)}}\\
= \dfrac{{1 - 8{{\cos }^2}x}}{{2 - 8{{\cos }^2}x}}
\end{array}\)
