If `A ={ theta : 2cos^2 theta + sintheta <=2}` , and `B = {theta: pi/2<=theta<= 3pi/2}` , then the region for `(AnnB)` is

Explanation

`because 2cos^(2)theta+sinthetale2`
`therefore 2(1-sin^(2)theta)+sinthetale2`
`implies 2sin^(2)theta-sinthetage0`
`implies sintheta(2sintheta-1)ge0`
`implies sintheta(sintheta-(1)/(2))ge0`
`therefore sinthetale0andsinthetage(1)/(2)`
Now, the values of `theta` which lie in teh interval `(pi)/(2)lethetale(3pi)/(2)[because B={theta:(pi)/(2)lethetale(3pi)/(2)}]`
So, `theta` satisfy `sin theta le 0` in the interval `(pi)/(2)lethetale(5pi)/(6)`.
`therefore AnnB={theta:pilethetale(3pi)/(2)}`
and `AnnB={theta:(pi)/(2)lethetale(5pi)/(6)}`
Hence, `AnnB={theta:(pi)/(2)lethetale(5pi)/(6)orpilethetale(3pi)/(2)}`
`={theta:thetain[(pi)/(2)(5pi)/(6)]uu[pi,(3pi)/(2)]}`