Giúp mình vs ạ. Mình cảm ơn nhiều 1.Cos(4x +60°)- 5cosx(2x + 30°)+ 4 = 0 2. Sin(2x+ 4π/3) + 2cos(x+ 2π/3) = 0

Đáp án:

$\begin{array}{l}
1)\cos \left( {4x + {{60}^0}} \right) – 5\cos \left( {2x + {{30}^0}} \right) + 4 = 0\\
 \Rightarrow \cos \left[ {2.\left( {2x + {{30}^0}} \right)} \right] – 5\cos \left( {2x + {{30}^0}} \right) + 4 = 0\\
 \Rightarrow 2{\cos ^2}\left( {2x + {{30}^0}} \right) – 1 – 5\cos \left( {2x + {{30}^0}} \right) + 4 = 0\\
 \Rightarrow 2{\cos ^2}\left( {2x + {{30}^0}} \right) – 5\cos \left( {2x + {{30}^0}} \right) + 3 = 0\\
 \Rightarrow \left( {\cos \left( {2x + {{30}^0}} \right) – 1} \right)\left( {2\cos \left( {2x + {{30}^0}} \right) – 3} \right) = 0\\
 \Rightarrow \cos \left( {2x + {{30}^0}} \right) = 1\\
 \Rightarrow 2x + {30^0} = k{360^0}\\
 \Rightarrow x =  – {15^0} + k{.180^0}\\
2)\\
\sin \left( {2x + \dfrac{{4\pi }}{3}} \right) + 2\cos \left( {x + \dfrac{{2\pi }}{3}} \right) = 0\\
 \Rightarrow \sin \left[ {2\left( {x + \dfrac{{2\pi }}{3}} \right)} \right] + 2\cos \left( {x + \dfrac{{2\pi }}{3}} \right) = 0\\
 \Rightarrow 2.\sin \left( {x + \dfrac{{2\pi }}{3}} \right).\cos \left( {x + \dfrac{{2\pi }}{3}} \right) + 2\cos \left( {x + \dfrac{{2\pi }}{3}} \right) = 0\\
 \Rightarrow 2\cos \left( {x + \dfrac{{2\pi }}{3}} \right)\left( {\sin \left( {x + \dfrac{{2\pi }}{3}} \right) + 1} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
\cos \left( {x + \dfrac{{2\pi }}{3}} \right) = 0\\
\sin \left( {x + \dfrac{{2\pi }}{3}} \right) =  – 1
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x + \dfrac{{2\pi }}{3} = \dfrac{\pi }{2} + k\pi \\
x + \dfrac{{2\pi }}{3} = \dfrac{{ – \pi }}{2} + k2\pi 
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{\pi }{6} + k\pi \\
x = \dfrac{{ – 7\pi }}{6} + k2\pi 
\end{array} \right.
\end{array}$