Câu 1:cos(3x-60°)= √3/2 Câu2:giải pt √ 3sinx -cosx = 1 Giúp mình mới ạ

Đáp án:

\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = 30^\circ  + k.120^\circ \\
x = 10^\circ  + k.120^\circ 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \pi  + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
\cos \left( {3x - 60^\circ } \right) = \dfrac{{\sqrt 3 }}{2}\\
 \Leftrightarrow \cos \left( {3x - 60^\circ } \right) = \cos 30^\circ \\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 60^\circ  = 30^\circ  + k.360^\circ \\
3x - 60^\circ  =  - 30^\circ  + k.360^\circ 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x = 90^\circ  + k.360^\circ \\
3x = 30^\circ  + k.360^\circ 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 30^\circ  + k.120^\circ \\
x = 10^\circ  + k.120^\circ 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)\\
2,\\
\sqrt 3 \sin x - \cos x = 1\\
 \Leftrightarrow \dfrac{{\sqrt 3 }}{2}\sin x - \dfrac{1}{2}\cos x = \dfrac{1}{2}\\
 \Leftrightarrow \sin x.\cos \dfrac{\pi }{6} - \cos x.\sin \dfrac{\pi }{6} = \dfrac{1}{2}\\
 \Leftrightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = \dfrac{1}{2}\\
 \Leftrightarrow \sin \left( {x - \dfrac{\pi }{6}} \right) = \sin \dfrac{\pi }{6}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{\pi }{6} = \dfrac{\pi }{6} + k2\pi \\
x - \dfrac{\pi }{6} = \dfrac{{5\pi }}{6} + k2\pi 
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{3} + k2\pi \\
x = \pi  + k2\pi 
\end{array} \right.\,\,\,\,\left( {k \in Z} \right)
\end{array}\)