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

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 = 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

4.08%

step-by-step explanation:

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.