Question

How do you solve `2tanA=-sec^2A`?

Answer 1

`{A|A = (3pi)/4 + npi, n in ZZ}`

Explanation:

It's possible to deal with a trigonometric equation using substitution if all terms use the same trigonometric function .

Starting with `2tanA = -sec^2 A`, we can use the identity:
`tan^2A + 1 = sec^2A`

to convert the `sec` term into a `tan` term.

`2tanA = -sec^2 A`
`2tanA = -(tan^2A+1)`

Then, we can rearrange the terms:

`tan^2A +2tanA+1=0`

This might look similar to a quadratic equation, and it can be solved in the same manner. The key step is to let a variable represent the trigonometric function, then solve for the values of that variable:

`tan^2A +2tanA+1=0`
Let `k = tanA`.
`k^2 + 2k + 1 = 0`
`(k+1)^2=0`

The solution to this equation is `k=-1`.

Then, we use our earlier substitution to solve for our original variable, `A`.

`k=-1`, `k = tanA`

`tan A = -1`

The tangent function has a period of `pi`, so let's first consider the solutions from `[0,pi]`. There is only one such solution:

`tan((3pi)/4)=-1`

Since the tangent function has a period of `pi`, the solution repeats every `pi` units.
Thus, the general solution is:
`{A|A = (3pi)/4 + npi, n in ZZ}`