If S = \ x ∈ ℝ: ^ - 1 ( x + 1 x ^ 2 + 2x + 2 ) - ^ - 1 ( x x ^ 2 +...

Given,

$\sin^{- 1}\left(\dfrac{x + 1}{\sqrt{{x}^{2} + 2x + 2}}\right) - \sin^{- 1}\left(\dfrac{x}{\sqrt{{x}^{2} + 1}}\right) = \dfrac{\pi}{4}$

$⇒ \sin^{- 1}\left(\dfrac{x + 1}{\sqrt{{x}^{2} + x + 2}}\right) = \dfrac{\pi}{4} + \sin^{- 1}\dfrac{x}{\sqrt{{x}^{2} + 1}}$

$⇒ \left(\dfrac{x + 1}{\sqrt{{x}^{2} + x + 2}}\right) = \sin \left(\dfrac{\pi}{4} + \sin^{- 1}\dfrac{x}{\sqrt{{x}^{2} + 1}}\right)$

$⇒ \left(\dfrac{x + 1}{\sqrt{{x}^{2} + x + 2}}\right) = \dfrac{1}{\sqrt{2}} \times \dfrac{1}{\sqrt{{x}^{2} + 1}} + \dfrac{1}{\sqrt{2}} \times \dfrac{x}{\sqrt{{x}^{2} + 1}}$

$⇒ \dfrac{x + 1}{\sqrt{{x}^{2} + x + 2}} = \dfrac{x + 1}{\sqrt{2}\sqrt{{x}^{2} + 1}}$

$⇒ \left(x + 1\right)\left(\sqrt{2}\sqrt{{x}^{2} + 1} - \sqrt{{x}^{2} + x + 2}\right) = 0$

$⇒ x = - 1$ or $\sqrt{{x}^{2} + x + 2} = \sqrt{2} \cdot \sqrt{{x}^{2} + 1}$

Now solving, $\sqrt{{x}^{2} + x + 2} = \sqrt{2} \cdot \sqrt{{x}^{2} + 1}$ we get,

${x}^{2} + x + 2 = 2\left({x}^{2} + 1\right)$

$⇒ {x}^{2} - x = 0$

$⇒ x = 0, x = 1$ {rejected as $x = 1$ will not satisfy the given equation}

Hence, $S = \left\{0, 1\right\}$

Now solving,

$\mathop{∑}_{n \in S}\left(\sin \left({x}^{2} + x + 5\right)\dfrac{\pi}{2} - \cos \left({x}^{2} + x + 5\right)\pi\right)$

$= \left(sin\dfrac{5\pi}{2} - \cos 5\pi + sin\dfrac{5\pi}{2} - \cos 5\pi\right)$

$= 1 - \left(- 1\right) + 1 - \left(- 1\right) = 4$