The number of solutions of secx−tanx=3,x∈[0,3π] is
4
3
2
1
We have secx−tanx=3⇒secx+tanx=13
Adding 2secx=3+13=43
⇒secx=23>0
Subtracting, we get 2tanx=13−3=−23
⇒tanx=−13<0
⇒x belongs to 4th quadrant
⇒x=11π6,23π6