If $A=\{θ:2 \cos^2 θ+ \sin θ ≤2\}$ and $B=\{θ:\frac{π}{2}≤θ ≤\frac{3π}{2}\}$ then AB is equal to

$\{θ:π/2≤θ≤5π/6\}∪\{θ:π≤θ≤3π/2\}$

We have,

$2 \cos^2θ+ \sin θ≤2$

$⇒2(1- \sin^2θ) + \sin θ ≤2$

$⇒2\sin^2θ-\sin θ≥0$

$⇒\sin θ (2 \sin θ-1) ≥0$

$⇒\sin θ≥0$ and $2\sin θ-1≥0$ or,

$⇒\sin θ≤0$ and $2\sin θ-1≤0$

CASE I When $\sin θ>0$ and $2\sin θ-1 >0$

In this case, we have

$\sin θ≥0$ and $2\sin θ-1≥0$

$⇒\sin θ≥0$ and $2\sin θ≥\frac{1}{2}$

$⇒\frac{π}{6}≤θ≤\frac{5π}{6}$

$∴A∩B=\{θ:π/2≤θ≤5π/6\}$   $[∵B=\{θ:π/2≤θ≤3π/2\}]$

CASE II When $\sin θ≤0$ and $2 \sin θ-1≤0$

In this case, we have

$⇒\sin θ≤0$ and $2\sin θ-1≤0$

$⇒\sin θ≤0$ and $2\sin θ≤\frac{1}{2}$

$⇒\sin θ≤0$

$⇒π≤θ≤2π$

$∴A∩B=\{θ:π≤θ≤3π/2\}$   $[∵B=\{θ:π/2≤θ≤3π/2\}]$

Thus, $A∩B=\{θ:π/2≤θ≤5π/6\}∪\{θ:π≤θ≤3π/2\}$