How do you solve the equation #3sqrt3tanu=3#?

1 Answer

#u=pi/6+2pix#
#u=(7pi)/6+2pix#
x is an integer

Explanation:

Some concepts needed when finding u:
#(x,y)=(cos,sin)#
#tan=sin/cos#

Math:
#3sqrt"3"tanu=3#
#tanu=3/(3sqrt3)#
#tanu=1/sqrt3#
#u=tan^-1(1/sqrt3)# and #tan^-1((-1)/-sqrt3)=pi/6 and (7pi)/6#
Unit Circle
This is because the ratio which fits one over square root of three is one half over two divided by square root of three since the twos cancel out. Negatives also cancel, which is why there are 2 answers.

In reality, there would be an infinite amount of solutions with equations since angles can keep going around. So since a full rotation is #2pi#, you can model the equation like
#u=pi/6+2pix# and #u=(7pi)/6+2pix# where x is an integer