题目详情 - Q20260425174444227

【分析】由二倍角公式结合题意可得答案. 【详解】$\sqrt{\dfrac{1+\cos\alpha}{2}}=\sqrt{\dfrac{1+2\cos^2\frac{\alpha}{2}-1}{2}}=\left|\cos\frac{\alpha}{2}\right|$， $\sqrt{\dfrac{1-\cos\alpha}{2}}=\sqrt{\dfrac{1-1+2\sin^2\frac{\alpha}{2}}{2}}=\left|\sin\frac{\alpha}{2}\right|$. 则$\left|\sin\frac{\alpha}{2}\right|+\left|\cos\frac{\alpha}{2}\right|=\sin\frac{\alpha}{2}-\cos\frac{\alpha}{2}\Rightarrow\begin{cases}\sin\frac{\alpha}{2}\ge0,\\\cos\frac{\alpha}{2}\le0\end{cases}$. 因$\alpha\in[0,2\pi]$，则$\frac{\alpha}{2}\in[0,\pi]$，又$\cos\frac{\alpha}{2}\le0$，则$\frac{\alpha}{2}\in\left[\frac{\pi}{2},\pi\right]\Rightarrow\alpha\in[\pi,2\pi]$. 故答案为：$[\pi,2\pi]$.