Phương trình lượng giác \(\dfrac{{\cos x - \dfrac{{\sqrt 3 }}{2}}}{{\sin x - \dfrac{1}{2}}} = 0\) có nghiệm là:

\(\left[ \begin{array}{l}x =  - \dfrac{\pi }{{36}} + \dfrac{{k2\pi }}{3}\\x = \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}\,\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

\(\left[ \begin{array}{l}x =  - \dfrac{\pi }{{36}} + \dfrac{{k\pi }}{3}\\x = \dfrac{{5\pi }}{{36}} + \dfrac{{k\pi }}{3}\,\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

\(\left[ \begin{array}{l}x = \dfrac{\pi }{{36}} + \dfrac{{k2\pi }}{3}\\x =  - \dfrac{{5\pi }}{{36}} + \dfrac{{k2\pi }}{3}\,\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

\(\left[ \begin{array}{l}x =  - \dfrac{\pi }{{36}} + k2\pi \\x = \dfrac{{5\pi }}{{36}} + k2\pi \,\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

$\min y = 1;\max y = 4033.$

\(\min y =  - 1;\max y = 4033.\)

\(\min y = 1;\max y = 4022.\)

\(\min y =  - 1;\max y = 4022.\)

\(1\)     

\(0\)      

\(3\)

\(2\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{8} + \dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{3\pi }}{8} + \dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{3\pi }}{8} + k\pi ,k \in \mathbb{Z}} \right\}\)

\(D = \mathbb{R}\backslash \left\{ {\dfrac{{3\pi }}{4} + \dfrac{{k\pi }}{2},k \in \mathbb{Z}} \right\}\) 

${\rm{D}} = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{3} + k2\pi ,\dfrac{{5\pi }}{3} + k2\pi \left| {k \in \mathbb{Z}} \right.} \right\}$

${\rm{D}} = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{3} + k2\pi \left| {k \in \mathbb{Z}} \right.} \right\}$

${\rm{D}} = \left\{ {\dfrac{\pi }{3} + k2\pi ,\dfrac{{5\pi }}{3} + k2\pi \left| {k \in \mathbb{Z}} \right.} \right\}$

${\rm{D}} = \mathbb{R}\backslash \left\{ {\dfrac{{5\pi }}{3} + k2\pi \left| {k \in \mathbb{Z}} \right.} \right\}$

\(\overrightarrow {MM'}  = \overrightarrow {NN'} \) 

\(\overrightarrow {MM'}  = \overrightarrow {N'N} \) 

\(\overrightarrow {MN'}  = \overrightarrow {NM'} \) 

\(\overrightarrow {MN}  = \overrightarrow u \)