Đáp án:
d) \(\tan x = - \dfrac{1}{{\sqrt 3 }}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)Do:x \in \left( {\pi ;\dfrac{{3\pi }}{2}} \right)\\
\to \left\{ \begin{array}{l}
\sin x < 0\\
\cos x < 0
\end{array} \right.\\
\tan x = \dfrac{3}{4}\\
\to \dfrac{{\sin x}}{{\cos x}} = \dfrac{3}{4}\\
\to \sin x = \dfrac{3}{4}\cos x\\
{\sin ^2}x + {\cos ^2}x = 1\\
\to \dfrac{9}{{16}}{\cos ^2}x + {\cos ^2}x = 1\\
\to {\cos ^2}x = \dfrac{{16}}{{25}}\\
\to \cos x = - \dfrac{4}{5}\\
\to \sin x = - \dfrac{3}{5}\\
\to \cot x = \dfrac{4}{3}\\
b)Do:x \in \left( {\dfrac{\pi }{2};\pi } \right)\\
\to \left\{ \begin{array}{l}
\cos x < 0\\
\sin x > 0
\end{array} \right.\\
\tan x = - \sqrt 2 \\
\to \sin x = - \sqrt 2 \cos x\\
{\sin ^2}x + {\cos ^2}x = 1\\
\to 2{\cos ^2}x + {\cos ^2}x = 1\\
\to {\cos ^2}x = \dfrac{1}{3}\\
\to \cos x = - \dfrac{1}{{\sqrt 3 }}\\
\to \sin x = \dfrac{{\sqrt 2 }}{{\sqrt 3 }}\\
\cot x = - \dfrac{1}{{\sqrt 2 }}\\
c)Do:x \in \left( {0;\dfrac{\pi }{2}} \right)\\
\to \left\{ \begin{array}{l}
\sin x > 0\\
\cos x > 0
\end{array} \right.\\
\cot x = \dfrac{2}{3}\\
\to \cos x = \dfrac{2}{3}\sin x\\
{\sin ^2}x + {\cos ^2}x = 1\\
\to {\sin ^2}x + \dfrac{4}{9}{\sin ^2}x = 1\\
\to {\sin ^2}x = \dfrac{9}{{13}}\\
\to \sin x = \dfrac{3}{{\sqrt {13} }}\\
\to \cos x = \dfrac{2}{{\sqrt {13} }}\\
\tan x = \dfrac{3}{2}\\
d)Do:x \in \left( {\dfrac{\pi }{2};\pi } \right)\\
\to \left\{ \begin{array}{l}
\cos x < 0\\
\sin x > 0
\end{array} \right.\\
\cot x = - \sqrt 3 \\
\to \cos x = - \sqrt 3 \sin x\\
{\sin ^2}x + {\cos ^2}x = 1\\
\to 3{\sin ^2}x + {\sin ^2}x = 1\\
\to {\sin ^2}x = \dfrac{1}{4}\\
\to \sin x = \dfrac{1}{2}\\
\to \cos x = - \dfrac{{\sqrt 3 }}{2}\\
\tan x = - \dfrac{1}{{\sqrt 3 }}
\end{array}\)
