The sum of all values of θ ∈ (0, (π)/(2)) satisfying ^ 2 2θ + ^ 4...

Given $\sin^{2}2\theta + \cos^{4}2\theta = \dfrac{3}{4}$

Using $\sin^{2}x + \cos^{2}x = 1,$ we get

$⇒ 1 - \cos^{2}2\theta + \cos^{4}2\theta = \dfrac{3}{4}$

Let $\cos^{2}2\theta = t$

$⇒ {t}^{2} - t + \dfrac{1}{4} = 0$

$⇒ {\left(t - \dfrac{1}{2}\right)}^{2} = 0$

$⇒ t = \dfrac{1}{2}$

$⇒ \cos^{2}2\theta = \dfrac{1}{2}$

$⇒ 2\cos^{2}2\theta - 1 = 0$

Using $\cos 2A = 2\cos^{2}A - 1,$ we get

$⇒ \cos 4\theta = 0$

If $\cos x = 0,$ then $x = \left(2n + 1\right)\dfrac{\pi}{2}, n \in Z .$

$⇒ 4\theta = \left(2n + 1\right)\dfrac{\pi}{2}$

$⇒ \theta = \left(2n + 1\right)\dfrac{\pi}{8}$

$⇒ \theta = \dfrac{\pi}{8}, \dfrac{3\pi}{8} \in \left(0, \dfrac{\pi}{2}\right)$

Hence, the sum of the values of $\theta$ is $\dfrac{\pi}{8} + \dfrac{3\pi}{8} = \dfrac{\pi}{2} .$