Let A = { theta : 2cos^2 theta + sin theta le | vedclass

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(D) Given $2\cos^2 	heta + \sin 	heta \le 2$. Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get $2(1 - \sin^2 	heta) + \sin 	heta \le 2$. $2 - 2\si

Vedclass · Accepted Answer

$\left\{ 	heta : 	heta \in \left[ \frac{\pi}{2}, \frac{5\pi}{6} ight] \cup \left[ \pi, \frac{3\pi}{2} ight] ight\}$

Vedclass · Answer

(D) Given $2\cos^2 	heta + \sin 	heta \le 2$. Using $\cos^2 	heta = 1 - \sin^2 	heta$,we get $2(1 - \sin^2 	heta) + \sin 	heta \le 2$. $2 - 2\sin^2 	heta + \sin 	heta \le 2$. $-2\sin^2 	heta + \sin 	heta \le 0$. Multiplying by $-1$,we get $2\sin^2 	heta - \sin 	heta \ge 0$. $\sin 	heta(2\sin 	heta - 1) \ge 0$. This inequality holds when $\sin 	heta \le 0$ or $\sin 	heta \ge \frac{1}{2}$. For $\sin 	heta \le 0$,$	heta \in [\pi, 2\pi]$. For $\sin 	heta \ge \frac{1}{2}$,$	heta \in [\frac{\pi}{6}, \frac{5\pi}{6}]$. Thus,$A = [\frac{\pi}{6}, \frac{5\pi}{6}] \cup [\pi, 2\pi]$. Given $B = [\frac{\pi}{2}, \frac{3\pi}{2}]$. Taking the intersection $A \cap B$: $A \cap B = ([\frac{\pi}{6}, \frac{5\pi}{6}] \cup [\pi, 2\pi]) \cap [\frac{\pi}{2}, \frac{3\pi}{2}]$. $A \cap B = [\frac{\pi}{2}, \frac{5\pi}{6}] \cup [\pi, \frac{3\pi}{2}]$.

Let $A = \{ \theta : 2\cos^2 \theta + \sin \theta \le 2 \}$ and $B = \{ \theta : \frac{\pi}{2} \le \theta \le \frac{3\pi}{2} \}$,then $A \cap B$ is

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