Đáp án:
$\begin{array}{l}
a)\sin 4x + \cos 4x = \sqrt 3 \\
\Leftrightarrow \sqrt 2 .\sin \left( {4x + \dfrac{\pi }{4}} \right) = \sqrt 3 \\
\Leftrightarrow \sin \left( {4x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 3 }}{{\sqrt 2 }}\\
Do:\left\{ \begin{array}{l}
- 1 \le \sin \left( {4x + \dfrac{\pi }{4}} \right) \le 1\\
\sqrt {\dfrac{3}{2}} > 1
\end{array} \right.
\end{array}$
Vậy pt vô nghiệm
$\begin{array}{l}
b)3\sin x + \sqrt 3 .\cos x = 1\\
\Leftrightarrow \dfrac{3}{{2\sqrt 3 }}\sin x + \dfrac{{\sqrt 3 }}{{2\sqrt 3 }}\cos x = \dfrac{1}{{2\sqrt 3 }}\\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin x + \dfrac{1}{2}\cos x = \dfrac{1}{{2\sqrt 3 }}\\
\Leftrightarrow \sin \dfrac{\pi }{3}.\sin x + \cos \dfrac{\pi }{3}.\cos x = \dfrac{1}{{2\sqrt 3 }}\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{3}} \right) = \dfrac{1}{{2\sqrt 3 }}\\
\Leftrightarrow x - \dfrac{\pi }{3} = \pm \arccos \dfrac{1}{{2\sqrt 3 }} + k2\pi \\
\Leftrightarrow x = \dfrac{\pi }{3} \pm \arccos \dfrac{1}{{2\sqrt 3 }} + k2\pi \\
Vậy\,x = \dfrac{\pi }{3} \pm \arccos \dfrac{1}{{2\sqrt 3 }} + k2\pi \\
\end{array}$
$\begin{array}{l}
c)\sqrt 3 \cos x + \sin x = - \sqrt 2 \\
\Leftrightarrow \dfrac{{\sqrt 3 }}{2}\cos x + \dfrac{1}{2}\sin x = - \dfrac{{\sqrt 2 }}{2}\\
\Leftrightarrow \cos \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{{ - \sqrt 2 }}{2}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = \dfrac{{3\pi }}{4} + k2\pi \\
x - \dfrac{\pi }{6} = - \dfrac{{3\pi }}{4} + k2\pi
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11\pi }}{{12}} + k2\pi \\
x = \dfrac{{ - 7\pi }}{{12}} + k2\pi
\end{array} \right.\\
Vậy\,\left[ \begin{array}{l}
x = \dfrac{{11\pi }}{{12}} + k2\pi \\
x = \dfrac{{ - 7\pi }}{{12}} + k2\pi
\end{array} \right.
\end{array}$
