# Can someone check my solution to a trigonometric equation?

## 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

This looks completely correct to me

Mark44
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.

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.

Mark44
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.

Oh, sorry.