Question

Let $A=\{\theta :2{\cos }^2\theta +\sin \theta \le 2\}$ and
$B=\{\theta :\pi /2\le \theta \le 3\pi /2\}$. Then find the value of$A\cap B$.

Answer 1

As given set $B$ consists of all values of $\theta$ in the interval $\pi /2\le \theta \le 3\pi /2$ and set $A$ consists of all values of $\theta$ which satisfy the inequality
$2{\cos }^2\theta +\sin \theta \le 2$.. . . $\left(1\right)$
Hence $A\cap B$ will consist of all those values of $\theta$ in the interval $\pi /2\le \theta \le 3\pi /2$ which satisfy the inequality $\left(1\right)$.
The inequality $\left(1\right)$ is equivalent to the inequality
$2-2{\sin }^2\theta +\sin \theta \le 2$,
that is, $\sin \theta (1-2\sin \theta )\le 0$
or $2\sin \theta (\sin \theta -\frac{1}{2})\ge 0$. . . $\left(2\right)$
The inequality $\left(2\right)$ is satisfied by all those values of $\theta$
which satisfy $\sin \theta \le 0$ or $\sin \theta \ge \frac{1}{2}$.
Now the values of $\theta$ which lie in the interval $\pi /2\le \theta \le 3\pi /2$ and satisfy $\sin \theta \le 0$ are given by $\pi \le \theta \le 3\pi /2$ And the values of $\theta$ which is in the interval $\pi /2\le \theta \le 3\pi /2$ and satisfy $\sin \theta \ge \frac{1}{2}$ are given by $\pi /2\le \theta \le 5\pi /6$.
Thus the solution set of inequality $\left(1\right)$ consists of all values of $\theta$ in the intervals $\pi /2\le \theta \le 5\pi /6$ and $\pi \le \theta \le 3\pi /2$. Hence
$A\cap B=\{\theta :\pi /2\le \theta \le 5\pi /6or\pi \le \theta \le 3\pi /2\}$

$A\cap B$=$[\theta :\frac{\pi }{2}\le \theta \le \frac{5\pi }{6}or\pi \le \theta \le \frac{3\pi }{2}]$