Đáp án:
\(\cot x = \dfrac{4}{3}\) hoặc \(\cot x = \dfrac{3}{4}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\sin x + \cos x = \dfrac{7}{5} \Leftrightarrow \cos x = \dfrac{7}{5} - \sin x\\
{\sin ^2}x + {\cos ^2}x = 1\\
\Leftrightarrow {\sin ^2}x + {\left( {\dfrac{7}{5} - \sin x} \right)^2} = 1\\
\Leftrightarrow {\sin ^2}x + \dfrac{{49}}{{25}} - \dfrac{{14}}{5}\sin x + {\sin ^2}x = 1\\
\Leftrightarrow 2{\sin ^2}x - \dfrac{{14}}{5}\sin x + \dfrac{{24}}{{25}} = 0\\
\Leftrightarrow 50{\sin ^2}x - 70\sin x + 24 = 0\\
\Leftrightarrow 25{\sin ^2}x - 35\sin x + 12 = 0\\
\Leftrightarrow \left( {5\sin x - 3} \right)\left( {5\sin x - 4} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sin x = \dfrac{3}{5} \Rightarrow \cos x = \dfrac{4}{5} \Rightarrow \cot x = \dfrac{{\cos x}}{{\sin x}} = \dfrac{4}{3}\\
\sin x = \dfrac{4}{5} \Rightarrow \cos x = \dfrac{3}{5} \Rightarrow \cot x = \dfrac{{\cos x}}{{\sin x}} = \dfrac{3}{4}
\end{array} \right.
\end{array}\)
Vậy \(\cot x = \dfrac{4}{3}\) hoặc \(\cot x = \dfrac{3}{4}\)
