If x,y∈0,15, then the number of solutions x,y of the equation  3cosec2x−1×4y2−4y+2≤1 is

We have 3cot2x2y−12+1≤1.But 3cot2x≥1and 2y−12≥1⇒3cot2x=1and 2y−12+1=1⇒cot2x=0,y=12⇒x=π2,3π2,5π2,7π2,9π2   ∵x∈0,15