Solve `sqrt(2) sec theta+tan theta=1`.

Explanation

To solve the equation \( \sqrt{2} \sec \theta + \tan \theta = 1 \), we can follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We know that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Substituting these into the equation gives:
\[ \sqrt{2} \cdot \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = 1 \]
This simplifies to:
\[ \frac{\sqrt{2} + \sin \theta}{\cos \theta} = 1 \]

### Step 2: Clear the fraction
Multiplying both sides by \( \cos \theta \) (assuming \( \cos \theta \neq 0 \)):
\[ \sqrt{2} + \sin \theta = \cos \theta \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ \cos \theta - \sin \theta = \sqrt{2} \]

### Step 4: Divide by \(\sqrt{2}\)
To simplify, we divide the entire equation by \(\sqrt{2}\):
\[ \frac{\cos \theta}{\sqrt{2}} - \frac{\sin \theta}{\sqrt{2}} = 1 \]
This can be rewritten using the cosine and sine of angles:
\[ \cos \left( \theta + \frac{\pi}{4} \right) = 1 \]

### Step 5: Solve for \(\theta\)
The general solution for \( \cos x = 1 \) is:
\[ x = 2n\pi \quad \text{for } n \in \mathbb{Z} \]
Thus, we have:
\[ \theta + \frac{\pi}{4} = 2n\pi \]
Solving for \(\theta\) gives:
\[ \theta = 2n\pi - \frac{\pi}{4} \]

### Final Solution
The general solution for the equation \( \sqrt{2} \sec \theta + \tan \theta = 1 \) is:
\[ \theta = 2n\pi - \frac{\pi}{4}, \quad n \in \mathbb{Z} \]