Khẳng định nào đúng về phương trình \(2\sqrt 2 \left( {\sin x + \cos x} \right)\cos x = 3 + \cos 2x\) 

\(\left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x =  \pm \dfrac{1}{2}\arccos \left( { - \dfrac{1}{6}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)\)     

\(\left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x =  \pm \dfrac{5}{2}\arccos \left( { - \dfrac{1}{6}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)\)

\(\left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x =  \pm \dfrac{1}{2}\arccos \left( { - \dfrac{1}{3}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)\)

\(\left[ \begin{array}{l}x = \dfrac{{k\pi }}{2}\\x =  \pm \dfrac{1}{3}\arccos \left( { - \dfrac{1}{6}} \right) + k\pi \end{array} \right.\left( {k \in Z} \right)\)

\(x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\)

\(x = \dfrac{\pi }{4} + \dfrac{{k\pi }}{2}\left( {k \in Z} \right)\)

\(x = \dfrac{{3\pi }}{4} + 2k\pi \left( {k \in Z} \right)\)        

\(x = \dfrac{{3\pi }}{4} + k\pi \left( {k \in Z} \right)\)

\(\left| a \right| \ge 1\)

\(\left| a \right| > 1\)   

\(\left| a \right| = 1\)

\(\left| a \right| \ne 1\)

\(\left\{ \begin{array}{l}x = \dfrac{\pi }{6} + k2\pi \\y =  - \dfrac{\pi }{6} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)          

\(\left\{ \begin{array}{l}x = \dfrac{{2\pi }}{3} + k2\pi \\y = \dfrac{\pi }{3} - k2\pi \end{array} \right.\left( {k \in Z} \right)\)                      

$\left\{ \begin{array}{l}x = \dfrac{{2\pi }}{3} + k2\pi \\y = \dfrac{\pi }{3} + k2\pi \end{array} \right.\left( {k \in Z} \right)$           

\(\left\{ \begin{array}{l}x = \dfrac{\pi }{2} + k2\pi \\y = \dfrac{\pi }{6} + k2\pi \end{array} \right.\left( {k \in Z} \right)\)

$\left[ \begin{array}{l}x = \dfrac{\pi }{3} + k\pi \\x = \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)$

$\left[ \begin{array}{l}x = \dfrac{\pi }{3} + k2\pi \\x = \dfrac{\pi }{6} + k2\pi \end{array} \right.\,\,\left( {k \in Z} \right)$

$\left[ \begin{array}{l}x =  - \dfrac{\pi }{3} + k\pi \\x =  - \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)$

$\left[ \begin{array}{l}x =  - \dfrac{\pi }{3} + k2\pi \\x = \dfrac{\pi }{6} + k\pi \end{array} \right.\,\,\left( {k \in Z} \right)$

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

\(x = \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)\)

\(x =  - \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)\)      

\(x =  \pm \dfrac{\pi }{6} + \dfrac{{k2\pi }}{3}\,\,\,\left( {k \in Z} \right)\)    

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

\(x =  \pm \dfrac{\pi }{6} + k2\pi \,\,\left( {k \in Z} \right)\)

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

\(\left[ \begin{array}{l}x = \dfrac{\pi }{6} + k\pi \,\,\left( {k \in Z} \right)\\x = \dfrac{\pi }{3} + k\pi \,\,\left( {k \in Z} \right)\end{array} \right.\)