Nghiệm của phương trình \(\cos 7x\cos 5x - \sqrt 3 \sin 2x = 1 - \sin 7x\sin 5x\) là:

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

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

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

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

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

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

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

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

\(x = 0\).

\(x = \pi \).

\(x = \dfrac{\pi }{3}\).

\(x = \dfrac{\pi }{2}\).

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

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

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

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

\(y = \sin x\)

\(y = \cos x\)

\(y = \sin 2x\)  

\(y = \cot x\) 

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

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

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

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

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

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

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

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