If tanx+secx=3, then the number of real solutions in [0,3π] is:

We are given: tan x + sec x = 3 We are to find the number of real solutions in the interval [0, 3π] . Let’s proceed with squaring both sides to eliminate tan and sec individually: Start with the identity: sec²x − tan²x = 1 This implies: sec x = √(1 + tan²x) But instead of introducing more complexity, try substituting tan x = t. Then, sec x = √(1 + t²) (since sec²x = 1 + tan²x) So the equation becomes: t + √(1 + t²) = 3 Now isolate the square root: √(1 + t²) = 3 − t Now square both sides: 1 + t² = (3 − t)² = 9 − 6t + t² Cancel t² from both sides: 1 = 9 − 6t → 6t = 8 → t = 4/3 So, tan x = 4/3 Now we need to find how many x satisfy tan x = 4/3 in [0, 3π] Since tan x = 4/3 has one solution in each period of π , and the interval [0, 3π] includes 3 full cycles , there will be 3 solutions .