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

 

[Mark44]

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.

 

[Mark44]

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.