Considering only the principal values of the inverse trigonometric...

Given,

$f\left(x\right) =$\cos^{- 1}\left(\dfrac{{x}^{2} - 4x + 2}{{x}^{2} + 3}\right)$</p><p>We know that domain of$\cos^{- 1}\text{is}\left[- 1, 1\right]

So, \left|\dfrac{{x}^{2} + 4x + 2}{{x}^{2} + 3}\right| \le 1$</p><p>$⇒ {\left({x}^{2} - 4x + 2\right)}^{2} \le {\left({x}^{2} + 3\right)}^{2}

$⇒ {\left({x}^{2} - 4x + 2\right)}$^{2} - {\left({x}^{2} + 3\right)}^{2} \le 0$</p><p>$⇒ \left(2{x}^{2} - 4x + 5\right)\left(- 4x - 1\right) \le 0$</p><p>Since \left(2{x}^{2} - 4x + 5\right)$ > 0$as its discriminant$D < 0$</p><p>So,$- 4x - 1 \le 0 ⇒ x \ge - $\dfrac{1}{4}</p><p>$⇒ x \in [ - $\dfrac{1}{4}$, \infty )$</p></blockquote>$