Question

Fill in the blanks :
A = $\theta :2co{s}^2\theta +sin\theta \le 2$
B = $\theta :\frac{\pi }{2}\le \frac{3\pi }{2}$ then A$\cap$B =........

Answer 1

$2co{s}^2\theta +sin\theta -2\le 0$
$\Rightarrow$ $2-2si{n}^2\theta +sin\theta -2\le 0$
or $2sin\theta (sin\theta -1/2)\ge 0$
or $2(sin\theta -0)(sin\theta -1/2)\ge 0$
$\therefore$ $sin\theta \le 0orsin\theta \ge 1/2.$
Now sin $\theta \le 0\Rightarrow \theta$ lies in 3rd or 4th quadrant for A, but in B, $\theta$ lies in 2nd and 3rd quadrant.
$\therefore$ A $\cap$ B = $\theta :\pi \le \theta \le 3\pi /2$, 3rd Q. ...(1)
Again, sin$\theta$ = 1 / 2
$\therefore$ $sin\theta \le 1/2if\pi /6\le \theta \le 5\pi /6$ for A but in B, $\theta$ lies in 2nd and 3rd quadrant
$\therefore$ $A\cap B$ = $\theta :\pi /2\le \theta \le 5\pi /6$, 2nd Q. ...(2)
$\therefore$ $A\cap B$ is given by (1) and (2).