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As stated in title

Guest Oct 18, 2017
 #1
avatar+349 
+1

First, note that \(\tan({\pi\over4})=1\). Imagine this on the unit circle. Since we need to find \(x\) when its tangent is negative 1, we have to get to the 2nd and 4th quadrant while retaining the tangent's absolute slope. To do that, we add \(90^o\) or \({\pi\over2}\) and \(270^o\) or \({3\pi\over2}\) to the angle \({\pi\over4}\). When we do that, we get the angles \({3\pi\over4}\) and \({7\pi\over4}\).

 

Q.E.D. wink

Mathhemathh  Oct 18, 2017
 #2
avatar+93866 
0

That is a good answer mathhemathh   laugh

but QED means 'it is proven' so it is only appropriate to use it at the end of proofs wink

Melody  Oct 18, 2017
 #4
avatar+349 
0

Malis verbis meis...

Mathhemathh  Oct 18, 2017
 #3
avatar+9677 
+1

tan(x) = -1        \(x=\frac{3}{4}\pi + k*\pi \)           \(k\epsilon Z \)

laugh

Omi67  Oct 18, 2017

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