Trig
posted by Kelly .
Finding all Solutions for
2cos(x)ã3=0

Is this cosx=1/2sqrt3 ?

no it's 2cosxsqrt3=0

I will read it as
2cosx  √3 = 0
cosx = √3/2
x = 30° , from the standard 306090 triangle
or x = 330°
"all solutions " would be
x = 30 + k(360) or x = 330+k(360)° where k is an integer. 
How do you find the solutions?

since the standard cosine curve has a period of 360°, any answer you have will repeat itself in the next curve or 360° either to the right or to the left.
by multiplying 360 by k, k and integer, I am simply adding/or subtracting multiples of 360 to any answer.
to get my original 30°, as I said, I recognized the rations of the 306090 rightangled triangle, which are 1:√3:2.
Secondly , since the cosine is positive in the 1st and 4th quadrant, the 30° will be the 1st quadrant solution and 36030 or 330° would be the 4th quadrant solution.