Answer:
The solutions of the equation are 0 , π
Explanation:
* Lets revise some trigonometric identities
- sin² Ф + cos² Ф = 1
- tan² Ф + 1 = sec² Ф
* Lets solve the equation
∵ tan² x sec² x + 2 sec² x - tan² x = 2
- Replace sec² x by tan² x + 1 in the equation
∴ tan² x (tan² x + 1) + 2(tan² x + 1) - tan² x = 2
∴ tan^4 x + tan² x + 2 tan² x + 2 - tan² x = 2 ⇒ add the like terms
∴ tan^4 x + 2 tan² x + 2 = 2 ⇒ subtract 2 from both sides
∴ tan^4 x + 2 tan² x = 0
- Factorize the binomial by taking tan² x as a common factor
∴ tan² x (tan² x + 2) = 0
∴ tan² x = 0
OR
∴ tan² x + 2 = 0
∵ 0 ≤ x < 2π
∵ tan² x = 0 ⇒ take √ for both sides
∴ tan x = 0
∵ tan 0 = 0 , tan π = 0
∴ x = 0
∴ x = π
OR
∵ tan² x + 2 = 0 ⇒ subtract 2 from both sides
∴ tan² x = -2 ⇒ no square root for negative value
∴ tan² x = -2 is refused
∴ The solutions of the equation are 0 , π