From the equation 2tanx - 1 = 0, we can find the value of tangent of x as follows:
2tanx - 1 = 0
2tanx = 1
tanx = 1/2
x = arctan(1/2) + nπ, where n is an integer
Now we need to solve √5cosx+4/sin2x, given that cos(x) < 0.
Since cos(x) < 0 and tan(x) = sin(x)/cos(x) = 1/2, we can use the Pythagorean identity to find sin(x):
sin^2(x) + cos^2(x) = 1
sin^2(x) + (-cos(x))^2 = 1
sin^2(x) = 1 - cos^2(x)
sin(x) = -√(1 - cos^2(x))
Substituting this into the expression we want to evaluate:
√5cos(x) + 4/sin^2(x) = √5cos(x) + 4/(-√(1 - cos^2(x)))^2
= √5cos(x) + 4/(1 - cos^2(x))
To simplify this expression further, we can use the identity sec^2(x) = 1 + tan^2(x) to find cos^2(x):
sec^2(x) = 1 + tan^2(x)
1/cos^2(x) = 1 + (1/2)^2
1/cos^2(x) = 5/4
cos^2(x) = 4/5
cos(x) = ±2/√5
Since cos(x) is negative, we have cos(x) = -2/√5.
Substituting this value into the expression we want to evaluate:
√5cos(x) + 4/(1 - cos^2(x)) = √5(-2/√5) + 4/(1 - (-2/√5)^2)
= -2 + 4/(1 - 4/5)
= -2 + 20
= 18
Therefore, the solution to √5cos(x) + 4/sin^2(x) given 2tan(x) - 1 = 0 and cos(x) < 0 is 18.