Nghiệm của phương trình ${\sin ^2}x + \sin x = 0$ thỏa điều kiện: $- \dfrac{\pi }{2} < x < \dfrac{\pi }{2}$.

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{4} + k\dfrac{\pi }{3},\dfrac{{n\pi }}{5};k,n \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{5} + k\dfrac{\pi }{3},\dfrac{{n\pi }}{5};k,n \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{6} + k\dfrac{\pi }{4},\dfrac{{n\pi }}{5};k,n \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{6} + k\dfrac{\pi }{3},\dfrac{{n\pi }}{5};k,n \in \mathbb{Z}} \right\}$

$x = \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)$

$x =  - \dfrac{\pi }{3} + k2\pi \left( {k \in Z} \right)$          

$x = \dfrac{\pi }{6} + k\pi \left( {k \in Z} \right)$

$x =  - \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)$

$k2\pi \left( {k \in Z} \right)$

$k2\pi ;\dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$

$\dfrac{{k\pi }}{2}\left( {k \in Z} \right)$

$k\pi ;\dfrac{\pi }{2} + k2\pi \left( {k \in Z} \right)$

$y = \sin x$

$y = \cos x$

$y = \sin 2x$  

$y = \cot x$ 

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{3} + k\dfrac{\pi }{2};k \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{4} + k\dfrac{\pi }{2};k \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{{12}} + k\dfrac{\pi }{2};k \in \mathbb{Z}} \right\}$

$D = \mathbb{R}\backslash \left\{ {\dfrac{\pi }{8} + k\dfrac{\pi }{2};k \in \mathbb{Z}} \right\}$

$y = \dfrac{\pi }{3} + k\pi \left( {k \in Z} \right)$

$x = k\pi \left( {k \in Z} \right)$

$x = \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)$

$y = \dfrac{{k\pi }}{2}\left( {k \in Z} \right)$