Answer:
The solutions of the equation is 0 , π , 2π
Explanation:
* Lets revise some identities in the trigonometry
- tan²x + 1 = sec²x
- tan²x = sec²x - 1
* Now lets solve the equation
∵ tan²x sec²x + 2sec²x - tan²x = 2
* Lets replace tan²x by sec²x - 1
∴ (sec²x - 1) sec²x + 2sec²x - (sec²x - 1) = 2 ⇒ open the brackets
∴ sec^4 x - sec²x + 2sec²x - sec²x + 1 = 2 ⇒ add the like terms
∴ sec^4 x -2sec²x + 2sec²x + 1 = 2 ⇒ cancel -2sec²x with +2sec²x
∴ sec^4 x + 1 = 2 ⇒ subtract 1 from both sides
∴ sec^4 x = 1 ⇒ take root four to both sides
∴ sec x = ± 1 ⇒ x is on the axes
∵ sec x = 1/cos x
∵ sec x = 1 , then cos x = 1
∵ sec x = -1 , then cos x = -1
∵ 0 ≤ x ≤ 2π
∵ x = cos^-1 (1)
∴ x = 0 , 2π ⇒ x is on the positive part of x-axis
∵ x = cos^-1 (-1)
∴ x = π ⇒ x is on the negative part of x-axis
* The solutions of the equation is 0 , π , 2π