BTVN - Phương trình lượng giác (Tiết 4) - Có lời g...

A \(\left[ \begin{array}{l}x = \frac{\pi }{3} + k\pi \\x = \arctan \left( {\frac{{ - \sqrt 3 }}{9}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)\)

B \(\left[ \begin{array}{l}x =  - \frac{\pi }{3} + k\pi \\x = \arctan \left( {\frac{{\sqrt 3 }}{9}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)\)

C \(x = \frac{\pi }{3} + k\pi \,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

D \(x =  - \frac{\pi }{3} + k2\pi \,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

A \(x = \frac{\pi }{2} + k\pi \), \(x = \arctan \frac{7}{{15}} + k\pi \left( {k \in Z} \right)\)

B \(x = \frac{\pi }{2} + k\pi \), \(x = \arctan \frac{3}{5} + k\pi \left( {k \in Z} \right)\)

C \(x = \frac{\pi }{2} + k\pi \), \(x = \arctan \frac{8}{{15}} + k\pi \left( {k \in Z} \right)\)

D \(x = \frac{\pi }{2} + k\pi \), \(x = \arctan \frac{2}{5} + k\pi \left( {k \in Z} \right)\)

A \(x = \frac{\pi }{6} + k2\pi \left( {k \in Z} \right)\)

B \(x = \frac{\pi }{6} + k\pi \left( {k \in Z} \right)\)

C \(x =  - \frac{\pi }{6} + k\pi \left( {k \in Z} \right)\)

D \(x =  - \frac{\pi }{6} + k2\pi \left( {k \in Z} \right)\)

A \(x = \arctan 3 + k\pi \)

B \(x = \arctan \left( { - 3} \right) + k2\pi \)

C \(x = \arctan \left( { - 3} \right) + k\pi \)

D \(x = \arctan 3 + k2\pi \)

A \(x = k\pi \), \(x =  \pm \frac{\pi }{4} + k2\pi \)

B \(x = k\pi \), \(x =  \pm \frac{\pi }{4} + k\pi \)

C \(x = k\pi \), \(x =  \pm \frac{\pi }{2} + k\pi \)

D \(x = k\pi \), \(x =  \pm \frac{\pi }{2} + k2\pi \)

A \(\left[ \begin{array}{l}x = \frac{\pi }{4} + k\pi \\x =  - \frac{\pi }{3} + k\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

B \(\left[ \begin{array}{l}x = \frac{\pi }{4} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

C \(\left[ \begin{array}{l}x =  \pm \frac{\pi }{4} + k2\pi \\x =  - \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\)

D \(\left[ \begin{array}{l}x =  \pm \frac{\pi }{4} + k\pi \\x =  - \frac{\pi }{3} + k\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\)