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Checkfate

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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 [tex]\Theta, 0\leq\Theta\leq2\pi[/tex]

[tex]2\tan\Theta-2=0[/tex]

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

This is how they show their work :

[tex]2\tan\Theta=2[/tex]

[tex]\tan\Theta=1[/tex]

[tex]\Theta=\frac{\pi}{4}[/tex] I don't fully grasp this step...

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

Thanks..

The question is : Solve for [tex]\Theta, 0\leq\Theta\leq2\pi[/tex]

[tex]2\tan\Theta-2=0[/tex]

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

This is how they show their work :

[tex]2\tan\Theta=2[/tex]

[tex]\tan\Theta=1[/tex]

[tex]\Theta=\frac{\pi}{4}[/tex] I don't fully grasp this step...

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

Thanks..

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