Đáp án:
$[_{x = \frac{\pi }{6} - \frac{{{\mathop{\rm ar}\nolimits} \cos \left( {\frac{{\sin \frac{\pi }{9}}}{2}} \right)}}{2} + k\pi }^{x = \frac{\pi }{6} + \frac{{{\mathop{\rm ar}\nolimits} \cos \left( {\frac{{\sin \frac{\pi }{9}}}{2}} \right)}}{2} + k\pi };k \in Z$
Giải thích các bước giải:
$\begin{array}{l}
2\cos (2x - \frac{\pi }{3}) - \sin {20^0} = 0\\
< = > \cos (2x - \frac{\pi }{3}) = \frac{{\sin \frac{\pi }{9}}}{2}\\
< = > [_{2x - \frac{\pi }{3} = - {\mathop{\rm ar}\nolimits} \cos \left( {\frac{{\sin \frac{\pi }{9}}}{2}} \right) + k2\pi }^{2x - \frac{\pi }{3} = {\mathop{\rm ar}\nolimits} \cos \left( {\frac{{\sin \frac{\pi }{9}}}{2}} \right) + k2\pi }\\
< = > [_{x = \frac{\pi }{6} - \frac{{{\mathop{\rm ar}\nolimits} \cos \left( {\frac{{\sin \frac{\pi }{9}}}{2}} \right)}}{2} + k\pi }^{x = \frac{\pi }{6} + \frac{{{\mathop{\rm ar}\nolimits} \cos \left( {\frac{{\sin \frac{\pi }{9}}}{2}} \right)}}{2} + k\pi };k \in Z
\end{array}$
