Solve each equation for x, where x is restricted to the given interval.
y = √2 + 3 sec 2x, for x in [0, π/4) ⋃ (π/4, π/2]

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Rewrite the given equation: $y = \sqrt{2} + 3 \sec 2x$. Our goal is to solve for $x$ in the interval $[0, \frac{\pi}{4}) \cup (\frac{\pi}{4}, \frac{\pi}{2}]$.

Isolate the $\sec 2x$ term by subtracting $\sqrt{2}$ from both sides: $y - \sqrt{2} = 3 \sec 2x$. Then divide both sides by 3 to get $\sec 2x = \frac{y - \sqrt{2}}{3}$.

Recall that $\sec \theta = \frac{1}{\cos \theta}$. So, rewrite the equation as $\frac{1}{\cos 2x} = \frac{y - \sqrt{2}}{3}$, which implies $\cos 2x = \frac{3}{y - \sqrt{2}}$.

Solve for \$2x$ by taking the inverse cosine (arccos) of both sides: $2x = \arccos \left( \frac{3}{y - \sqrt{2}} \right)$. Remember that cosine is positive in the first and fourth quadrants, so consider all possible solutions for \$2x$ within the domain.

Finally, divide by 2 to solve for $x$: $x = \frac{1}{2} \arccos \left( \frac{3}{y - \sqrt{2}} \right)$. Check which solutions fall within the given interval $[0, \frac{\pi}{4}) \cup (\frac{\pi}{4}, \frac{\pi}{2}]$ and exclude any values where $\sec 2x$ is undefined (such as at $x = \frac{\pi}{4}$).

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