# Can someone check my solution to a trigonometric equation?

Eclair_de_XII

## Homework Statement

4sec2θ tanθ = 8 tanθ

## Homework Equations

sec2θ = tan2θ + 1

## The Attempt at a Solution

4(tan2 + 1) tanθ = 8 tanθ
(4tan2 + 4) tanθ = 8 tanθ
4 tan3 + 4 tanθ = 8 tanθ
4 tan3 - 4 tanθ = 0
4tanθ(tan2θ - 1) = 0

4tanθ = 0
tanθ = 0
tan-1(tanθ) = tan-1(0)
tan-1θ = 0, π

tan2θ - 1 = 0
(tanθ + 1)(tanθ - 1) = 0

tanθ + 1 = 0
tanθ = -1
tan-1tanθ = tan-1(-1)
tan-1θ = 3π/4, 7π/4

tanθ - 1 = 0
tanθ = 1
tan-1tanθ = tan-1(1)
tan-1θ = π/4, 5π/4

hdp12
This looks completely correct to me

Mentor

## Homework Statement

4sec2θ tanθ = 8 tanθ

## Homework Equations

sec2θ = tan2θ + 1

## The Attempt at a Solution

4(tan2 + 1) tanθ = 8 tanθ
(4tan2 + 4) tanθ = 8 tanθ
4 tan3 + 4 tanθ = 8 tanθ
4 tan3 - 4 tanθ = 0
4tanθ(tan2θ - 1) = 0

4tanθ = 0
tanθ = 0
tan-1(tanθ) = tan-1(0)
tan-1θ = 0, π

tan2θ - 1 = 0
(tanθ + 1)(tanθ - 1) = 0

tanθ + 1 = 0
tanθ = -1
tan-1tanθ = tan-1(-1)
tan-1θ = 3π/4, 7π/4

tanθ - 1 = 0
tanθ = 1
tan-1tanθ = tan-1(1)
tan-1θ = π/4, 5π/4
Your solutions above are fine if the question is asking for solutions θ such that 0 ≤ θ ≤ 2π, but if the question is asking for all solutions, then the work above is not complete.

I should also add that some of what you wrote is incorrect.
tan-1θ = 0, π
This should be θ = 0 or θ = π -- the tan-1 business shouldn't be there. The same is true on the other two sections where you have done this.

Eclair_de_XII
but if the question is asking for all solutions, then the work above is not complete.

Isn't it just θ = 3π/4 + kπ or θ = π/4 + kπ?

This should be θ = 0 or θ = π -- the tan-1 business shouldn't be there. The same is true on the other two sections where you have done this.

Okay, that's useful to know.

Mentor
Isn't it just θ = 3π/4 + kπ or θ = π/4 + kπ?
Yes, but since you didn't include any information about the expected solutions, I didn't know exactly what the problem was asking for.

Eclair_de_XII
Oh, sorry.