If 3tanA.tanA-4√3TanA+3=0

TanA=n,3n.n-4√3n+3=0,

Are you solving 3tanA.tanA-4√3TanA+3=0 ?

What does the . in 3tanA.tanA mean? Is it multiplication?

I have no clue what
TanA=n,3n.n-4√3n+3=0
in the second part of your post means.

To solve the quadratic equation 3tan(A)·tan(A) - 4√3tan(A) + 3 = 0, we can use the quadratic formula. The quadratic formula states that for an equation in the form ax^2 + bx + c = 0, the solutions for x are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = 3, b = -4√3, and c = 3. Plugging these values into the quadratic formula, we have:

tan(A) = (-(-4√3) ± √((-4√3)^2 - 4·3·3)) / (2·3)

Simplifying this expression further, we get:

tan(A) = (4√3 ± √(48 - 36)) / 6

tan(A) = (4√3 ± √12) / 6

tan(A) = (√3 ± √3) / 3

Now, we have two possible solutions for tan(A):

1) tan(A) = (√3 + √3) / 3
Simplifying this expression further:
tan(A) = (2√3) / 3

2) tan(A) = (√3 - √3) / 3
Simplifying this expression further:
tan(A) = 0

Therefore, the solutions for A are:
1) A = atan((2√3) / 3)
2) A = atan(0)

To determine the exact values of A, you can use a calculator or refer to a table of trigonometric ratios.