Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
5,\\
\sin \left( {3x + 45^\circ } \right) - \cos \left( {x + 120^\circ } \right) = 0\\
\Leftrightarrow \cos \left( {x + 120^\circ } \right) = \sin \left( {3x + 45^\circ } \right)\\
\Leftrightarrow \cos \left( {x + 120^\circ } \right) = \cos \left[ {90^\circ - \left( {3x + 45^\circ } \right)} \right]\\
\Leftrightarrow \cos \left( {x + 120^\circ } \right) = \cos \left( {45^\circ - 3x} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
x + 120^\circ = 45^\circ - 3x + k.360^\circ \\
x + 120^\circ = 3x - 45^\circ + k.360^\circ
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - \dfrac{{75^\circ }}{4} + k.90^\circ \\
x = \dfrac{{165^\circ }}{2} + k.180^\circ
\end{array} \right.\,\,\,\,\,\,\,\,\,\,\left( {k \in Z} \right)\\
6,\\
\cos \left( {x + \dfrac{\pi }{3}} \right) + \cos \left( {x - \dfrac{\pi }{3}} \right) = 1\\
\Leftrightarrow 2.\cos \dfrac{{\left( {x + \dfrac{\pi }{3}} \right) + \left( {x - \dfrac{\pi }{3}} \right)}}{2}.\cos \dfrac{{\left( {x + \dfrac{\pi }{3}} \right) - \left( {x - \dfrac{\pi }{3}} \right)}}{2} = 1\\
\Leftrightarrow 2.\cos x.\cos \dfrac{\pi }{3} = 1\\
\Leftrightarrow \cos x = 1\\
\Leftrightarrow x = k2\pi \,\,\,\,\left( {k \in Z} \right)
\end{array}\)
