4(cosx)^2-3=0 ????????

4(cosx)^2 -3= 0
4 (cosx)^2 = 3
(cosx)^2 = 3/4

cos x = ( sq rt 3)/2

x = 30 degrees

4(cosx)^2-3=0
e^(-2 i x)+e^(2 i x)-1 = 0
x = pi n-(5 pi)/6, n element Z
x = pi n-pi/6, n element Z

4cos²(x) - 3 = 0
cos²(x) = 3/4
cos(x) = ±√3/2

Principal values of x are
x = {π/6, 5π/6}

4cos^2(x) - 3 = 0

Isolate the cos^2(x).

4cos^2(x) = 3
cos^2(x) = 3/4

Take the square root of both sides (don't forget the plus/minus!)

cos(x) = +/- sqrt(3/4)

When rooting a fraction, we root the numerator and denominator.

cos(x) = +/- sqrt(3)/sqrt(4)
cos(x) = +/- sqrt(3)/2

Fortunately for us, both sqrt(3)/2 and -sqrt(3)/2 are known values on the unit circle. Split into two equations, solve for each.

cos(x) = sqrt(3)/2
cos(x) = -sqrt(3)/2

Solve each.

x = { pi/6, 11pi/6 }
x = { 5pi/6, 7pi/6 }

Combine solutions.

x = {pi/6, 5pi/6, 7pi/6, 11pi/6 }

cos ² x = 3 / 4

cos x = ± √3 / 2

x = 30 ° , 150 ° , 210 ° , 330 °
OR
x = π / 6 , 5 π / 6 , 7 π / 6 , 10 π / 6

4[cos(x)]^2 - 3 = 0
4[cos(x)]^2 = 3
[cos(x)]^2 = 3/4
cos(x) = ±√(3/4)
cos(x) = ±(√3)/2
x = 30°, 150°, 210° and 330°