Question

Find all solutions ​

Answer 1

Answer:

$x= 0$  and $x =\pi$

Step-by-step 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

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 = 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