Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
\sin x + \sqrt 3 \cos x = 1\\
 \Leftrightarrow \dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}.\cos x = \dfrac{1}{2}\\
 \Leftrightarrow \sin x.\cos \dfrac{\pi }{3} + \cos x.\sin \dfrac{\pi }{3} = \dfrac{1}{2}\\
 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{3}} \right) = \sin \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{\pi }{3} = \dfrac{\pi }{6} + k2\pi \\
x + \dfrac{\pi }{3} = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{\pi }{2} + k2\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
b,\\
\sqrt 3 .\sin x + \cos x = 2\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}.\sin x + \dfrac{1}{2}\cos x = 1\\
 \Leftrightarrow \sin x.\cos \dfrac{\pi }{6} + \cos x.\sin \dfrac{\pi }{6} = 1\\
 \Leftrightarrow \sin \left( {x + \dfrac{\pi }{6}} \right) = 1\\
 \Leftrightarrow x + \dfrac{\pi }{6} = \dfrac{\pi }{2} + k2\pi \\
 \Leftrightarrow x = \dfrac{\pi }{3} + k2\pi \,\,\,\,\,\left( {k \in Z} \right)\\
c,\\
\sin x - \sqrt 3 \cos x =  - 1\\
 \Leftrightarrow \dfrac{1}{2}\sin x - \dfrac{{\sqrt 3 }}{2}.\cos x =  - \dfrac{1}{2}\\
 \Leftrightarrow \sin x.\cos \dfrac{\pi }{3} - \cos x.\sin \dfrac{\pi }{3} =  - \dfrac{1}{2}\\
 \Leftrightarrow \sin \left( {x - \dfrac{\pi }{3}} \right) = \sin \left( { - \dfrac{\pi }{6}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{3} =  - \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{3} = \dfrac{{7\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k2\pi \\
x = \dfrac{{3\pi }}{2} + k2\pi 
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
d,\\
\sin 3x - \cos 3x = \sqrt 2 \\
 \Leftrightarrow \dfrac{{\sqrt 2 }}{2}.\sin 3x - \dfrac{{\sqrt 2 }}{2}\cos 3x = 1\\
 \Leftrightarrow \sin 3x.\cos \dfrac{\pi }{4} - \cos 3x.\sin \dfrac{\pi }{4} = 1\\
 \Leftrightarrow \sin \left( {3x - \dfrac{\pi }{4}} \right) = 1\\
 \Leftrightarrow 3x - \dfrac{\pi }{4} = \dfrac{\pi }{2} + k2\pi \\
 \Leftrightarrow 3x = \dfrac{{3\pi }}{4} + k2\pi \\
 \Leftrightarrow x = \dfrac{\pi }{4} + \dfrac{{k2\pi }}{3}\,\,\,\,\,\left( {k \in Z} \right)\\
e,\\
\cos 7x - \sin 5x = \sqrt 3 .\left( {\cos 5x - \sin 7x} \right)\\
 \Leftrightarrow \cos 7x + \sqrt 3 \sin 7x = \sqrt 3 \cos 5x + \sin 5x\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}.\sin 7x + \dfrac{1}{2}\cos 7x = \dfrac{1}{2}\sin 5x + \dfrac{{\sqrt 3 }}{2}\cos 5x\\
 \Leftrightarrow \sin 7x.\cos \dfrac{\pi }{6} + \cos 7x.\sin \dfrac{\pi }{6} = \sin 5x.\cos \dfrac{\pi }{3} + \cos 5x.\sin \dfrac{\pi }{3}\\
 \Leftrightarrow \sin \left( {7x + \dfrac{\pi }{6}} \right) = \sin \left( {5x + \dfrac{\pi }{3}} \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
7x + \dfrac{\pi }{6} = 5x + \dfrac{\pi }{3} + k2\pi \\
7x + \dfrac{\pi }{6} = \dfrac{{2\pi }}{3} - 5x + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{{12}} + k\pi \\
x = \dfrac{\pi }{{24}} + \dfrac{{k\pi }}{6}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
f,\\
{\sin ^4}x - {\cos ^4}x = 1 + 2\sqrt 3 .\sin x.\cos x\\
 \Leftrightarrow \left( {{{\sin }^2}x - {{\cos }^2}x} \right).\left( {{{\sin }^2}x + {{\cos }^2}x} \right) = 1 + 2\sqrt 3 .\sin x.\cos x\\
 \Leftrightarrow \left( {{{\sin }^2}x - {{\cos }^2}x} \right).1 = {\sin ^2}x + {\cos ^2} + 2\sqrt 3 .\sin x.\cos x\\
 \Leftrightarrow {\sin ^2}x - {\cos ^2}x = {\sin ^2}x + {\cos ^2}x + 2\sqrt 3 .\sin x.\cos x\\
 \Leftrightarrow 2{\cos ^2}x + 2\sqrt 3 \sin x.\cos x = 0\\
 \Leftrightarrow \cos x.\left( {\cos x + \sqrt 3 \sin x} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
\cos x = 0\\
\cos x + \sqrt 3 \sin x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\dfrac{1}{2}\cos x + \dfrac{{\sqrt 3 }}{2}\sin x = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sin x.\cos \dfrac{\pi }{6} + \cos x.\sin \dfrac{\pi }{6} = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
\sin \left( {x + \dfrac{\pi }{6}} \right) = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x + \dfrac{\pi }{6} = k\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{2} + k\pi \\
x =  - \dfrac{\pi }{6} + k\pi 
\end{array} \right.\,\,\,\,\,\,\left( {k \in Z} \right)\\
g,\\
4{\cos ^3}x - \sqrt 3 \sin 3x = 1 + 3\cos x\\
 \Leftrightarrow \left( {4{{\cos }^3}x - 3\cos x} \right) - \sqrt 3 \sin 3x = 1\\
 \Leftrightarrow \cos 3x - \sqrt 3 \sin 3x = 1\\
 \Leftrightarrow \dfrac{1}{2}\cos 3x - \dfrac{{\sqrt 3 }}{2}.\sin 3x = \dfrac{1}{2}\\
 \Leftrightarrow \cos 3x.\cos \dfrac{\pi }{3} - \sin 3x.\sin \dfrac{\pi }{3} = \dfrac{1}{2}\\
 \Leftrightarrow \cos \left( {3x + \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow \left[ \begin{array}{l}
3x + \dfrac{\pi }{3} = \dfrac{\pi }{3} + k2\pi \\
3x + \dfrac{\pi }{3} =  - \dfrac{\pi }{3} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{k2\pi }}{3}\\
x =  - \dfrac{{2\pi }}{9} + \dfrac{{k2\pi }}{3}
\end{array} \right.\,\,\,\,\,\left( {k \in Z} \right)\\
h,\\
{\cos ^3}x - 3\sqrt 3 {\sin ^3}x = \sqrt 3 \sin x - \cos x\\
 \Leftrightarrow {\cos ^3}x - {\left( {\sqrt 3 \sin x} \right)^3} = \sqrt 3 \sin x - \cos x\\
 \Leftrightarrow \left( {\cos x - \sqrt 3 \sin x} \right).\left( {{{\cos }^2}x + \cos x.\sqrt 3 \sin x + 3{{\sin }^2}x} \right) = \sqrt 3 \sin x - \cos x\\
 \Leftrightarrow \left( {\cos x - \sqrt 3 \sin x} \right).\left[ {\left( {{{\cos }^2}x + \sqrt 3 \sin x.\cos x + 3{{\sin }^2}x} \right) + 1} \right] = 0\\
 \Leftrightarrow \left( {\cos x - \sqrt 3 \sin x} \right).\left[ {{{\cos }^2}x + \sqrt 3 \sin x.\cos x + 3{{\sin }^2}x + \left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right] = 0\\
 \Leftrightarrow \left( {\cos x - \sqrt 3 \sin x} \right).\left( {2{{\cos }^2}x + \sqrt 3 \sin x.\cos x + 4{{\sin }^2}x} \right) = 0\\
4{\sin ^2}x + \sqrt 3 \sin x.\cos x + 2{\cos ^2}x\\
 = {\left( {2\sin x} \right)^2} + 2.2\sin x.\dfrac{{\sqrt 3 }}{4}\cos x + \dfrac{3}{{16}}{\cos ^2}x + \dfrac{{29}}{{16}}{\cos ^2}x\\
 = {\left( {2\sin x + \dfrac{{\sqrt 3 }}{4}\cos x} \right)^2} + \dfrac{{29}}{{16}}{\cos ^2}x > 0,\,\,\,\forall x\\
 \Rightarrow \cos x - \sqrt 3 \sin x = 0\\
 \Leftrightarrow \dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x = 0\\
 \Leftrightarrow \cos x.\cos \dfrac{\pi }{3} - \sin x.\sin \dfrac{\pi }{3} = 0\\
 \Leftrightarrow \cos \left( {x + \dfrac{\pi }{3}} \right) = 0\\
 \Leftrightarrow x + \dfrac{\pi }{3} = \dfrac{\pi }{2} + k\pi \\
 \Leftrightarrow x = \dfrac{\pi }{6} + k\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)