trigonometry quadratic- I'm unsure how to get past the 4sqrt3

Question

4sqrt(3)cos (x)-4cos^2 (x)-3=0

 

Pi <x <2pi

Answer 1

It is nothing else but a quadratic equation:

 

- 4cos²(x) +4√3cos(x) - 3 =0

 

multiply by -1 both sides to yield

 

4cos²(x) -4√3cos(x) +3 =0

 

Now let z = cos(x) and you get the equation

 

4z² - 4√3z +3 = 0         (How do you solve the equation ax²+bx +c = 0  where a=4, b= -4√3, and c = 3 ?)

 

This equation can be solved and one finds that it has two coincident roots:  z = √3/2.

 

However, we are not after z, but x.  To find x, now we solve the trigonometric equation:

 

cox(x) = √3/2   which implies that x = Pi/6 (or 30 degrees) and  x = 11/6* Pi (or 330 degrees).

 

Finally, you are told that Pi <x <2Pi and thus the only solution is  x = 11/6* Pi.