Question

How do you solve `tanx=tan^2x` in the interval `0<=x<=2pi`?

Answer 1

`x = 0, (pi) / (4), (3 pi) / (4), pi, (5 pi) / (4), (7 pi) / (4), 2 pi`

Explanation:

We have: `tan(x) = tan^(2)(x)`; `0 leq x leq 2 pi`

First, let's subtract `tan(x)` from both sides of the equation:

`=> tan^(2)(x) - tan(x) = 0`

Then, let's take `tan(x)` in common:

`=> tan(x) (tan(x) - 1) = 0`

Now that we have a product that is equal to zero, either one of the multiples must be equal to zero:

`=> tan(x) = 0`

`=> x = 0, (pi - 0), (pi + 0), (2 pi - 0)`

`=> x = 0, pi, 2 pi`

or

`=> tan(x) - 1 = 0`

`=> tan(x) = 1`

`=> x = (pi) / (4), (pi - (pi) / (4)), (pi + (pi) / (4)), (2 pi - (pi) / (4))`

`=> x = (pi) / (4), (3 pi) / (4), (5 pi) / (4), (7 pi) / (4)`

Therefore, the solutions to the equation are `x = 0, (pi) / (4), (3 pi) / (4), pi, (5 pi) / (4), (7 pi) / (4), 2 pi`.