Sorry there are quite a few problems, but I just need to know if these are correct (and if they aren't, where I went wrong):

1. Solve these equations.
tanθ = -√3

θ = 2π/3 + kπ
θ = 5π/3 + kπ

2. secθ = √2

θ = π/4 + 2kπ
θ = 7π/4 + 2kπ

3. tanθ = -√3/3

θ = 5π/6 + kπ
θ = 11π/6 + kπ

4. cscθ = √2

θ = π/4 + 2kπ
θ = 3π/4 + 2kπ

Use a calculator to solve these, on the interval 0 ≤ θ ≤ 2π:

1. cotθ = -5
θ = 3.338988213
θ = 6.480580867

I know this is wrong. The answers are 2.94 and 6.09 (rounded). But shouldn't the answers be from π-θ and 2π-θ? Theta is negative in the given equation, so shouldn't it be π+θ and 2π+θ?

2. sinθ = 0.432
θ = 0.4456092078
θ = 2.694883446

3. cotθ = -2

Same deal with this problem. shouldn't it be π+θ and 2π+θ because theta is negative?? Or have I memorized the wrong formula for each quadrant?
These are the correct answers:
θ = 2.6779
θ = 5.8195

Thank you so much!

The first 4 are all ok

cot is negative in QII and QIV, and you are right about π±θ. But, that θ is the reference angle, which is always in QI. θ = arccot(5), not -5.

Let's go through each problem to see where you might have made a mistake.

1. Solve these equations.
tanθ = -√3

Your answers for this question are correct. The general solution for tanθ = -√3 is:
θ = 2π/3 + kπ
θ = 5π/3 + kπ

2. secθ = √2

Again, your answers for this question are correct. The general solution for secθ = √2 is:
θ = π/4 + 2kπ
θ = 7π/4 + 2kπ

3. tanθ = -√3/3

Your answers for this question are also correct. The general solution for tanθ = -√3/3 is:
θ = 5π/6 + kπ
θ = 11π/6 + kπ

4. cscθ = √2

Once more, your answers for this question are correct. The general solution for cscθ = √2 is:
θ = π/4 + 2kπ
θ = 3π/4 + 2kπ

Now, let's move on to the calculator problems.

1. cotθ = -5

The solution should indeed be π+θ and 2π+θ because cotθ is negative. So the correct answers are:
θ ≈ 2.94
θ ≈ 6.09

You are correct in questioning the method. The solutions should be π+θ and 2π+θ for negative cotθ.

2. sinθ = 0.432

Your answers for this question are correct. There is no need to use π+θ or 2π+θ for this problem.

3. cotθ = -2

Your answers for this question are correct. Again, there is no need to use π+θ or 2π+θ for this problem.

It seems like you might have misunderstood when to use the π+θ or 2π+θ adjustments. In general, you only need to use those adjustments when the trigonometric function is negative. When the function is positive, you can simply rely on the general solutions without any adjustments.

I hope this explanation clears up any confusion. Feel free to ask if you have any more questions!