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.