To solve the trigonometric equation

$\sqrt{2} \sec x + 2 = 0$

for all values $0 \leq x < 2\pi$, we proceed as follows:

Solution Steps:

1. Isolate $\sec x$: $\sqrt{2} \sec x = -2$ $\sec x = -\frac{2}{\sqrt{2}}$ Simplifying: $\sec x = -\sqrt{2}$

2. Recall that $\sec x = \frac{1}{\cos x}$. Thus: $\frac{1}{\cos x} = -\sqrt{2}$ Therefore: $\cos x = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$

3. Identify the angles where $\cos x = -\frac{\sqrt{2}}{2}$. This occurs at: $x = \frac{3\pi}{4} \quad \text{and} \quad x = \frac{5\pi}{4}$ within the interval $0 \leq x < 2\pi$.

Final Answer:

The solutions to the equation in the interval $0 \leq x < 2\pi$ are: $x = \frac{3\pi}{4}, \quad x = \frac{5\pi}{4}$

Would you like more detailed explanations on any steps or have any questions? Here are some related questions to further explore:

1. What are the properties of secant and how does it relate to cosine?
2. How do we convert between secant values and cosine values?
3. What are the trigonometric values of other common angles such as $\pi/4$ and $3\pi/2$?
4. What is the significance of the interval $0 \leq x < 2\pi$ in trigonometric problems?
5. How would this problem change if $\sec x$ was positive?

Tip: Remember that $\sec x$ is undefined when $\cos x = 0$, so watch out for domain restrictions in similar problems.