Question

Find all solutions ​

Answer 1

`x= 0` and
`x =\pi`

Explanation:

Remember the following trigonometric property

`tan^2(x) + 1 = sec^2(x)`

We have the following equation

`tan^2(x)*sec^2(x)+2sec^2(x)-tan^2(x) =2` for
`0\leq x <2\pi`

Using the mentioned property we have to:

`tan^2(x)*(tan^2(x) + 1)+2(tan^2(x) + 1)-tan^2(x) =2`

`tan^2(x)*(tan^2(x) + 1)+2tan^2(x) + 2-tan^2(x) =2`

`tan^2(x)*(tan^2(x) + 1)+2tan^2(x) -tan^2(x) =0`

`tan^2(x)*(tan^2(x) + 1)+tan^2(x) =0`

Take
`tan^2(x)` as a common factor

`tan^2(x)*[(tan^2(x) + 1)+1] =0`

`tan^2(x)*(tan^2(x) + 2) =0`

Then:

`tan^2(x)= 0` or
`tan^2(x)+2=0` →
`tan^2(x)=-2`

`tan(x) = 0` when
`x= 0` and
`x =\pi`

`tan^2(x)=-2` there is no solution for this case

Finally the solutions are:

`x= 0` and
`x =\pi`

Answer 2

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