Đáp án:
\(\tan x = - \dfrac{1}{{\sqrt 6 }};\sin x = \dfrac{1}{{\sqrt 7 }};\cos x = - \sqrt {\dfrac{6}{7}} \)
Giải thích các bước giải:
\(\begin{array}{l}
- \dfrac{{5\pi }}{2} < x < - \dfrac{{3\pi }}{2}\\
\to - 5\pi < 2x < - 3\pi \\
\to - 2\pi - 3\pi < 2x < - 2\pi - \pi \\
\to - 3\pi < 2x < - \pi \\
\to - \dfrac{{3\pi }}{2} < x < - \dfrac{\pi }{2}\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\cos x < 0\\
\sin x > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
\cos x < 0\\
\sin x < 0
\end{array} \right.
\end{array} \right.\\
TH1:\left\{ \begin{array}{l}
\cos x < 0\\
\sin x > 0
\end{array} \right.\\
\cot x = - \sqrt 6 \\
\to \dfrac{{\cos x}}{{\sin x}} = - \sqrt 6 \\
\to \cos x = - \sqrt 6 \sin x\\
{\cos ^2}x + {\sin ^2}x = 1\\
\to 6{\sin ^2}x + {\sin ^2}x = 1\\
\to {\sin ^2}x = \dfrac{1}{7}\\
\to \sin x = \dfrac{1}{{\sqrt 7 }} \to \cos x = - \sqrt {\dfrac{6}{7}} \\
\to \tan x = - \dfrac{1}{{\sqrt 6 }}\\
TH2:\left\{ \begin{array}{l}
\cos x < 0\\
\sin x < 0
\end{array} \right.\\
\cot x = - \sqrt 6 \\
\to \dfrac{{\cos x}}{{\sin x}} = - \sqrt 6 \\
\to \cos x = - \sqrt 6 \sin x\\
{\cos ^2}x + {\sin ^2}x = 1\\
\to 6{\sin ^2}x + {\sin ^2}x = 1\\
\to {\sin ^2}x = \dfrac{1}{7}\\
\to \sin x = - \dfrac{1}{{\sqrt 7 }} \to \cos x = \sqrt {\dfrac{6}{7}} \left( l \right)\\
\to TH2\left( l \right)\\
KL:\tan x = - \dfrac{1}{{\sqrt 6 }};\sin x = \dfrac{1}{{\sqrt 7 }};\cos x = - \sqrt {\dfrac{6}{7}}
\end{array}\)
