If 4 sin² θ = 3(1 + cos θ), 0° < θ < 90°, then what is the value of (2 tan θ + 4 sin θ - sec θ)?

To find the value of $2 \tan \theta +4 \sin \theta -\sec \theta$, we must first solve the given trigonometric equation for $\theta$ in the interval $0°<\theta <90°$:

$4{\sin }^2\theta =3(1+\cos \theta )$

Using the fundamental trigonometric identity ${\sin }^2\theta =1-{\cos }^2\theta$, we can substitute this into the equation to express everything in terms of $\cos \theta$:
$4(1-{\cos }^2\theta )=3(1+\cos \theta )$

Expand and rewrite the equation:
$4-4{\cos }^2\theta =3+3\cos \theta$

Rearranging all terms to one side gives a quadratic equation in terms of $\cos \theta$:
$4{\cos }^2\theta +3\cos \theta -1=0$

We can factor this quadratic equation by splitting the middle term:
$4{\cos }^2\theta +4\cos \theta -\cos \theta -1=0$
$4\cos \theta (\cos \theta +1)-1(\cos \theta +1)=0$
$(4\cos \theta -1)(\cos \theta +1)=0$

This yields two possible values for $\cos \theta$:
1) $\cos \theta =-1$
2) $\cos \theta =\frac{1}{4}$

Since we are given that $0°<\theta <90°$, the cosine of $\theta$ must be positive (first quadrant). Therefore, we discard $\cos \theta =-1$ and accept:
$\cos \theta =\frac{1}{4}$

Using $\cos \theta =\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{1}{4}$, we can determine the opposite side using the Pythagorean theorem:
$\text{Opposite}=\sqrt{4^2-1^2}=\sqrt{16-1}=\sqrt{15}$

Now, we find the values of the required trigonometric functions:
$\sin \theta =\frac{\sqrt{15}}{4}$
$\tan \theta =\frac{\sqrt{15}}{1}=\sqrt{15}$
$\sec \theta =\frac{4}{1}=4$

Substitute these values into the target expression $2\tan \theta +4\sin \theta -\sec \theta$:
$2\left(\sqrt{15}\right)+4\left(\frac{\sqrt{15}}{4}\right)-4$
$=2\sqrt{15}+\sqrt{15}-4$
$=3\sqrt{15}-4$

Thus, the final value of the expression is $3\sqrt{15}-4$.