每日一题[2750]复合逻辑

已知 $\alpha \in \mathbb R$，如果集合 $\{\sin \alpha,\cos {2\alpha}\}=\{\cos \alpha,\sin{2\alpha}\}$，则所有符合要求的角 $\alpha$ 构成的集合为_______．

答案    $\{\alpha\mid \alpha=2k\pi,k\in \mathbb Z\}$．

解析    根据题意，有 $$(\sin\alpha\ne \cos2\alpha)\land \left(\begin{cases} \sin\alpha=\cos\alpha,\\ \cos2\alpha=\sin2\alpha,\end{cases}\lor \begin{cases} \sin\alpha=\sin2\alpha,\\ \cos2\alpha=\cos\alpha,\end{cases}\right)$$ 即 $$\left(\sin\alpha\ne \dfrac 12,-1\right)\land\left( \begin{cases} \sqrt 2\sin\left(\alpha-\dfrac{\pi}4\right)=0,\\ \sqrt2\sin\left(2\alpha-\dfrac{\pi}4\right)=0,\end{cases}\lor \begin{cases} \sin\alpha(2\cos\alpha-1)=0,\\ \cos\alpha=1,-\dfrac 12,\end{cases}\right),$$ 也即 $$\left(\sin\alpha\ne \dfrac 12,-1\right)\land\left( \begin{cases} \alpha=\dfrac{\pi}4+k_1\pi,\\ \alpha=\dfrac{\pi}8+\dfrac{k_2\pi}2,\end{cases}\lor \begin{cases} \sin\alpha=0,\\ \cos\alpha=1,\end{cases}\right),$$ 其中 $k_1,k_2\in\mathbb Z$，解得 $\alpha$ 的范围是 $\{\alpha\mid \alpha=2k\pi,k\in \mathbb Z\}$．