# Let A= theta : 2cos ^2 theta +sin theta less than equal to 2 and B = theta : pi/2 less than equal to ( theta) less than equal to 3pi/2 Find A intersection B=?

Solution:

$\\A=\{\theta :2-2\sin^{2}\theta +\sin \theta \leq 2\}\\ \\=\{\theta :\sin \theta (1-2\sin \theta )\leq 0\}$

Above is possible if  $\sin \theta \leq 0$   or  $\sin \theta \geq \frac{1}{2}$

$\sin \theta =\frac{1}{2}$   when $\theta =\frac{\pi }{6}$  and is greater than half $\frac{1}{2}$

When    $\frac{\pi }{6}\leq \theta \leq \frac{5\pi }{6}$

Also    $\sin \theta \leq 0$  where $\pi \leq \theta \leq 2\pi .$

$\therefore$     $A=\{\theta :\frac{\pi }{6}\leq \frac{5\pi }{6} \hspace{0.2cm}or \hspace{0.2cm}\pi \leq \theta \leq 2\pi \}$

$B=\{\theta :\frac{\pi }{2}\leq \theta \leq \frac{3\pi }{2}\}$

$\therefore$    $A\cap B=\{\theta :\frac{\pi}{2}\leq \theta \leq \frac{5\pi}{6} \hspace{0.2cm}or \hspace{0.2cm}\pi\leq \theta \leq 2\pi\}$

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