
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

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system

Given that \( \sin x \neq 0 \) and \( \cos x \neq 0 \), multiply both sides of the equation by \( \operatorname{tg} x \). We obtain \( \sqrt{1+\operatorname{tg} x} = \sin x + \cos x \).

Under the condition \( \sin x + \cos x \geq 0 \), both sides of this equation can be squared. Since \( \sin x + \cos x = \sqrt{2} \sin \left(x + \frac{\pi}{4}\right) \), the inequality holds if \( -\frac{\pi}{4} + 2\pi n \leq x \leq \frac{3\pi}{4} + 2\pi n \) for \( n \in \mathbb{Z} \). Within the given constraints and conditions \( \sin x \neq 0 \) and \( \cos x \neq 0 \), the equation becomes equivalent to the following:

\[ 1+\operatorname{tg} x = \sin^2 x + 2 \sin x \cos x + \cos^2 x \quad \Leftrightarrow \quad \operatorname{tg} x - 2 \sin x \cos x = 0 \quad \Leftrightarrow \quad \frac{\sin x}{\cos x} - 2 \sin x \cos x = 0 \]

This simplifies to:

\[ \sin x \left(\frac{1}{\cos x} - 2 \cos x \right) = 0 \]

Thus, we have the equation:

\[ \cos^2 x = \frac{1}{2} \]

So,

\[ \cos x = \pm \frac{\sqrt{2}}{2} \]
 
Hence, 

\[ x = \pm \frac{\pi}{4} + \pi n, \quad n \in \mathbb{Z} \]

Considering the constraints, the solutions to the original equation are:

\[ x = \pm \frac{\pi}{4} + 2\pi n, \quad x = \frac{3\pi}{4} + 2\pi n, \quad n \in \mathbb{Z} \]

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