Solving for Theta: Understanding the Last Step

[Checkfate]

I don't really understand a step in my textbook, and I think it is better that I understand why they are doing it this way instead of just brushing it off and it coming back to haunt me later!

The question is : Solve for $$\Theta, 0\leq\Theta\leq2\pi$$

$$2\tan\Theta-2=0$$

I get the right answer, but their last step confuses me...

This is how they show their work :

$$2\tan\Theta=2$$

$$\tan\Theta=1$$

$$\Theta=\frac{\pi}{4}$$ I don't fully grasp this step...

I understand that tan is 1 at $$\frac{\pi}{4}$$ since sin and cosine are both $$\frac{\sqrt{3}}{2}$$ at that point. But how can you call tan pi/2 when it also is 1 at $$\frac{5\pi}{4}$$... Can someone explain what they are doing? They don't even give $$\frac{5\pi}{4}$$ as an answer in their equation, they just state it later in a sentence after the question.

Thanks..

 

[Checkfate]

Nevermind I understand now, fairly simple. Just convert the equation to $$\Theta=tan^{-1}(1)$$

$$\Theta=\frac{\pi}{4}$$

This trigonometry stuff is slowly eating my brain, lol. ahh!

 

[homology]

I guess I don't understand. $\frac{5\pi}{4}$ and $\frac{\pi}{4}$ both work, they are both solutions to the equation $2tan\theta -2 = 0$. Reducing the equation to $\theta=tan^{-1}(1)$ isn't the answer since you have to pick which part of tangent you want to invert. Putting $tan^{-1}(1)$ into your calculator will give you $\frac{\pi}{4}$ but this is not the only answer. It would be if your inequality read something like $0\leq \theta \leq \frac{\pi}{2}$ but it doesn't. You're looking for all solutions between zero and $2\pi$. I'm not sure why your book doesn't include both solutions though, that's odd. But you were right to be bothered by this.

 

[semc]

hmm...$$\frac{\pi}{4}$$ is your alpha.after getting your alpha,you are suppose to use it in the general solution and for tan,it is n$${pi}$$+$$\frac{\pi}{4}$$ where n is all real number.since the question ask for $$\Theta, 0\leq\Theta\leq2\pi$$, the only 2 are $$\frac{5\pi}{4}$$ and $$\frac{\pi}{4}$$ coz your n can only take 0 and 1.get it?

 

[Checkfate]

I guess that makes sense I just thought that they should show that in their algebraic answer, but what you are saying makes sense and seems to coincide with their instructions. Thanks