Solving Trigonometric Equations

[eutas1]

Please refer to attachment A

 

[Steven G]

-pi/4 + n*pi = 3pi/4 + (n-1)*pi !!!!

That is the two solutions are the same

 

[jonah2.0]

Beer soaked observation follows.

Please refer to attachment A

It could be just the beer but I'm pretty sure that should have been:
Tangent is negative in the second and fourth quadrants.

 

[eutas1]

-pi/4 + n*pi = 3pi/4 + (n-1)*pi !!!!

That is the two solutions are the same

I don't understand where -(pi/4) + n*pi comes from?
And where did you get 3pi/4 + (n-1)*pi ? Why is it (n-1) and not just 'n' ?
I thought tan's cycle was only pi so why are they considering 2pi ? Shouldn't you only consider quadrant 1 and 2??

 

[eutas1]

Beer soaked observation follows.

View attachment 26330
View attachment 26331
It could be just the beer but I'm pretty sure that should have been:
Tangent is negative in the second and fourth quadrants.

Ah, this would make sense! However, tan's cycle is pi, not 2pi? So I don't understand why they are even considering ALL 4 quadrants - shouldn't it just be quadrant 1 and 2?