Đáp án:
\(\begin{array}{l}
a)\quad\left[\begin{array}{l}x = \dfrac{2\pi}{3} + k2\pi\\x = - \dfrac{2\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
b)\quad\left[\begin{array}{l}x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{\pi}{2} +k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
c)\quad\left[\begin{array}{l}x = \dfrac{7\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\quad \sin\left(\dfrac{\pi}{2} - x\right) = -\sin\dfrac{\pi}{6}\\
\Leftrightarrow \cos x = -\dfrac12\\
\Leftrightarrow \cos x = \cos\dfrac{2\pi}{3}\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{2\pi}{3} + k2\pi\\x = - \dfrac{2\pi}{3} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có họ nghiệm là}\ x = \pm \dfrac{2\pi}{3} + k2\pi\ \text{với}\ k\in\Bbb Z\\
b)\quad \cos\left(x - \dfrac{\pi}{3}\right) = \cos\left(-\dfrac{\pi}{6}\right)\\
\Leftrightarrow \left[\begin{array}{l}x - \dfrac{\pi}{3} = -\dfrac{\pi}{6} + k2\pi\\x - \dfrac{\pi}{3} = \dfrac{\pi}{6} +k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{\pi}{6} + k2\pi\\x = \dfrac{\pi}{2} +k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có họ nghiệm là}\ x = \dfrac{\pi}{6} + k2\pi\ \text{và}\ x = = \dfrac{\pi}{2} +k2\pi\ \text{với}\ k\in\Bbb Z\\
c)\quad\sin\left(x - \dfrac{\pi}{6}\right) = \sin(\pi - 2x)\\
\Leftrightarrow \left[\begin{array}{l}x - \dfrac{\pi}{6} = \pi - 2x + k2\pi\\x - \dfrac{\pi}{6} = 2x + k2\pi\end{array}\right.\\
\Leftrightarrow \left[\begin{array}{l}x = \dfrac{7\pi}{18} + k\dfrac{2\pi}{3}\\x = \dfrac{\pi}{6} + k2\pi\end{array}\right.\quad (k\in\Bbb Z)\\
\text{Vậy phương trình có họ nghiệm là}\ x = \dfrac{7\pi}{18} + k\dfrac{2\pi}{3}\ \text{và}\ x = \dfrac{\pi}{6} + k2\pi\ \text{với}\ k\in\Bbb Z
\end{array}\)
