General solution to tan

[VCE_2012]

Correct me if I'm wrong:
Find the general solution to tan(x)=root(3)

after some working

x=pi/3+2*pi*k, (4*pi)/3+2*pi*k,k=Z

[daniel034]

isn't it just pi*k + pi/3 so you get -2pi/3, pi/3, 4pi/3, 7pi/3 etc

[BubbleWrapMan]

Remember that the period of , n > 0 is rather than like and

[VCE_2012]

isn't it just pi*k + pi/3 so you get -2pi/3, pi/3, 4pi/3, 7pi/3 etc

yes, but when I sub k=0 I get the solutions (pi/3 and 4pi/3) which lie between the (0,2pi)
and when I sub in k=1 I get solution which lie between the next revolution.

[BubbleWrapMan]

Well yes that works but it can be condensed into one expression

[VCE_2012]

I think it should be

I've seen

Well yes that works but it can be condensed into one expression

Yeh, I am used to it so I'll probably stick to it comes the exam, as long it's mathematically correct.