Find all solutions of each equation on the interval 0≤x<2 pi.tan^2x sec^2 x+2 sec^2x -tan^2x=2 Show your work

Answer:The solutions of the equation are 0 , π Step-by-step 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 = πOR∵ 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