The sum of all values of $$\theta \in (0, \frac{\pi}{2})$$ satisfying $$\sin^2 2\theta + \cos^4 2\theta = \frac{3}{4}$$ is:

We need all angles $$\theta$$ lying in the open interval $$(0,\dfrac{\pi}{2})$$ that satisfy the trigonometric equation

$$\sin^2 2\theta+\cos^4 2\theta=\dfrac34.$$

First, to simplify the expression, we put

$$x=2\theta.$$

Because $$\theta\in(0,\dfrac{\pi}{2})$$, doubling every number in this interval gives

$$x\in(0,\pi).$$

Under this substitution the equation becomes

$$\sin^2 x+\cos^4 x=\dfrac34.$$

Next, we use the Pythagorean identity $$\sin^2 x+\cos^2 x=1.$$ Solving this for $$\sin^2 x$$ gives

$$\sin^2 x=1-\cos^2 x.$$

Substituting this in the transformed equation produces

$$\bigl(1-\cos^2 x\bigr)+\cos^4 x=\dfrac34.$$

Now let us introduce another temporary variable to avoid writing $$\cos$$ repeatedly. Put

$$y=\cos^2 x.$$

Because $$x\in(0,\pi),$$ the value of $$\cos x$$ ranges from ±1 down to 0, so $$y=\cos^2 x$$ must satisfy $$0<y<1.$$

Under this change of variable, the equation simplifies to a quadratic:

$$1-y+y^2=\dfrac34.$$

We move every term to the left so that the right-hand side becomes zero:

$$1-y+y^2-\dfrac34=0.$$

Combining the constant terms on the left, we have

$$y^2-y+\bigl(1-\dfrac34\bigr)=0,$$

so

$$y^2-y+\dfrac14=0.$$

Observe that the left side is a perfect square. Indeed,

$$(y-\dfrac12)^2=y^2-y+\dfrac14.$$

Therefore we can rewrite the quadratic as

$$(y-\dfrac12)^2=0.$$

Solving this gives a single real root

$$y=\dfrac12.$$

But $$y=\cos^2 x,$$ so

$$\cos^2 x=\dfrac12.$$

Taking square roots on both sides, we obtain

$$\cos x=\pm\dfrac1{\sqrt2}.$$

Now we must find every $$x$$ in the interval $$(0,\pi)$$ that satisfies this relation.

We know from basic trigonometry that

$$\cos x=\dfrac1{\sqrt2}\quad\text{at}\quad x=\dfrac{\pi}{4},$$

and

$$\cos x=-\dfrac1{\sqrt2}\quad\text{at}\quad x=\dfrac{3\pi}{4}.$$

Both of these angles indeed lie strictly between $$0$$ and $$\pi,$$ so the corresponding values of $$x$$ are

$$x_1=\dfrac{\pi}{4},\qquad x_2=\dfrac{3\pi}{4}.$$

We now reverse the substitution $$x=2\theta$$ to back-substitute for $$\theta$$:

For $$x_1$$:

$$2\theta_1=\dfrac{\pi}{4}\;\Longrightarrow\;\theta_1=\dfrac{\pi}{8}.$$

For $$x_2$$:

$$2\theta_2=\dfrac{3\pi}{4}\;\Longrightarrow\;\theta_2=\dfrac{3\pi}{8}.$$

Both values $$\dfrac{\pi}{8}$$ and $$\dfrac{3\pi}{8}$$ lie inside the required interval $$(0,\dfrac{\pi}{2}),$$ so they are the only solutions for $$\theta.$$

The question asks for the sum of all such $$\theta.$$ Adding the two values we get

$$\theta_1+\theta_2=\dfrac{\pi}{8}+\dfrac{3\pi}{8}=\dfrac{4\pi}{8}=\dfrac{\pi}{2}.$$

Hence, the correct answer is Option A.