Giải dum em nha!!!!!!!!!!!

Đáp án:

 b) $x = \dfrac{\pi }{8} + \dfrac{1}{2}\arccos \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right) + k\pi $; $x = \dfrac{\pi }{8} - \dfrac{1}{2}\arccos \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right) + k\pi $;$x = \dfrac{\pi }{4} + k\pi $;$x = k\pi $$\k \in Z$

c) $x = k2\pi $;$x =  - \dfrac{\pi }{2} + k2\pi $;$x =  - \dfrac{\pi }{4} + \arccos \left( {\dfrac{{ - 2}}{{3\sqrt 2 }}} \right) + k2\pi $;$x =  - \dfrac{\pi }{4} - \arccos \left( {\dfrac{{ - 2}}{{3\sqrt 2 }}} \right) + k2\pi $$k\in Z$

d) $x = \dfrac{{3\pi }}{8} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$

Giải thích các bước giải:

$\begin{array}{l}
b)\sin 2x + \cos 2x - 4\sin 2x.\cos 2x - 1 = 0\left( 1 \right)\\
t = \sin 2x + \cos 2x\left( {\left| t \right| \le \sqrt 2 } \right)\\
 \Rightarrow {t^2} = 1 + 2.\sin 2x.\cos 2x\\
 \Rightarrow \sin 2x.\cos 2x = \dfrac{{{t^2} - 1}}{2}\\
\left( 1 \right)tt:t - 4.\dfrac{{{t^2} - 1}}{2} - 1 = 0\\
 \Leftrightarrow  - 2{t^2} + t + 1 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
t = \dfrac{{ - 1}}{2}\\
t = 1
\end{array} \right.\\
 + )t = \dfrac{{ - 1}}{2}\\
 \Rightarrow \sin 2x + \cos 2x = \dfrac{{ - 1}}{2}\\
 \Leftrightarrow \dfrac{1}{{\sqrt 2 }}\left( {\sin 2x + \cos 2x} \right) = \dfrac{{ - 1}}{{2\sqrt 2 }}\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) = \dfrac{{ - 1}}{{2\sqrt 2 }}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{4} = \arccos \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right) + k2\pi \\
2x - \dfrac{\pi }{4} =  - \arccos \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right) + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{8} + \dfrac{1}{2}\arccos \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right) + k\pi \\
x = \dfrac{\pi }{8} - \dfrac{1}{2}\arccos \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right) + k\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 + )t = 1\\
 \Rightarrow \sin 2x + \cos 2x = 1\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
2x - \dfrac{\pi }{4} =  - \dfrac{\pi }{4} + k2\pi 
\end{array} \right.\left( {k \in Z} \right) \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{4} + k\pi \\
x = k\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$

Vậy phương trình có các họ nghiệm là: $x = \dfrac{\pi }{8} + \dfrac{1}{2}\arccos \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right) + k\pi $; $x = \dfrac{\pi }{8} - \dfrac{1}{2}\arccos \left( {\dfrac{{ - 1}}{{2\sqrt 2 }}} \right) + k\pi $;$x = \dfrac{\pi }{4} + k\pi $;$x = k\pi $$\k \in Z$

$\begin{array}{l}
c)\cos x - \sin x + 3\sin 2x - 1 = 0\left( 2 \right)\\
t = \cos x - \sin x\left( {\left| t \right| \le \sqrt 2 } \right)\\
 \Rightarrow {t^2} = 1 - 2\sin x.\cos x = 1 - \sin 2x\\
 \Rightarrow \sin 2x = 1 - {t^2}\\
\left( 2 \right)tt:t + 3\left( {1 - {t^2}} \right) - 1 = 0\\
 \Leftrightarrow  - 3{t^2} + t + 2 = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
t = 1\\
t = \dfrac{{ - 2}}{3}
\end{array} \right.\\
 + )t = 1\\
 \Rightarrow \cos x - \sin x = 1\\
 \Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\
x + \dfrac{\pi }{4} =  - \dfrac{\pi }{4} + k2\pi 
\end{array} \right.\left( {k \in Z} \right) \Leftrightarrow \left[ \begin{array}{l}
x = k2\pi \\
x =  - \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 + )t = \dfrac{{ - 2}}{3}\\
 \Rightarrow \cos x - \sin x = \dfrac{{ - 2}}{3}\\
 \Leftrightarrow \cos \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{ - 2}}{{3\sqrt 2 }}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{4} = \arccos \left( {\dfrac{{ - 2}}{{3\sqrt 2 }}} \right) + k2\pi \\
x + \dfrac{\pi }{4} =  - \arccos \left( {\dfrac{{ - 2}}{{3\sqrt 2 }}} \right) + k2\pi 
\end{array} \right.\left( {k \in Z} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{4} + \arccos \left( {\dfrac{{ - 2}}{{3\sqrt 2 }}} \right) + k2\pi \\
x =  - \dfrac{\pi }{4} - \arccos \left( {\dfrac{{ - 2}}{{3\sqrt 2 }}} \right) + k2\pi 
\end{array} \right.\left( {k \in Z} \right)
\end{array}$

Vậy phương trình có cá họ nghiệm là: $x = k2\pi $;$x =  - \dfrac{\pi }{2} + k2\pi $;$x =  - \dfrac{\pi }{4} + \arccos \left( {\dfrac{{ - 2}}{{3\sqrt 2 }}} \right) + k2\pi $;$x =  - \dfrac{\pi }{4} - \arccos \left( {\dfrac{{ - 2}}{{3\sqrt 2 }}} \right) + k2\pi $$k\in Z$

$\begin{array}{l}
d)\sin 2x.\cos 2x + \sqrt 2 \left( {\cos 2x - \sin 2x} \right) = \dfrac{1}{2}\left( 3 \right)\\
t = \cos 2x - \sin 2x\left( {\left| t \right| \le \sqrt 2 } \right)\\
 \Rightarrow {t^2} = 1 - 2.\sin 2x.\cos 2x\\
 \Rightarrow \sin 2x.\cos 2x = \dfrac{{1 - {t^2}}}{2}\\
\left( 3 \right)tt:\dfrac{{1 - {t^2}}}{2} + \sqrt 2 .t = \dfrac{1}{2}\\
 \Leftrightarrow  - {t^2} + 2\sqrt 2 .t = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
t = 0\left( c \right)\\
t = 2\sqrt 2 \left( l \right)
\end{array} \right.\\
 \Rightarrow t = 0\\
 \Rightarrow \cos 2x - \sin 2x = 0\\
 \Leftrightarrow \cos \left( {2x - \dfrac{\pi }{4}} \right) = 0\\
 \Leftrightarrow 2x - \dfrac{\pi }{4} = \dfrac{\pi }{2} + k\pi \left( {k \in Z} \right)\\
 \Leftrightarrow x = \dfrac{{3\pi }}{8} + k\dfrac{\pi }{2}\left( {k \in Z} \right)
\end{array}$

Vậy phương trình có họ nghiệm là: $x = \dfrac{{3\pi }}{8} + k\dfrac{\pi }{2}\left( {k \in Z} \right)$