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

1 Answer
Sep 7, 2015

{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}