The most general value of `theta` which satisfy both the equation `cos theta =-(1)/(sqrt(2)) and tan theta =1 ` , is

Options

A. `2n pi + (5pi)/(4), n in Z`
B. `2n pi + (pi)/(4), n in Z`
C. `2 n pi + (3 pi)/(4), n in Z`
D. none of these

Answer

To find the most general value of \( \theta \) that satisfies both equations \( \cos \theta = -\frac{1}{\sqrt{2}} \) and \( \tan \theta = 1 \), we will follow these steps:

### Step 1: Analyze \( \cos \theta = -\frac{1}{\sqrt{2}} \)

The cosine function is negative in the second and third quadrants. The reference angle for \( \cos \theta = \frac{1}{\sqrt{2}} \) is \( \frac{\pi}{4} \). Therefore, the angles in the second and third quadrants where \( \cos \theta = -\frac{1}{\sqrt{2}} \) can be expressed as:
- In the second quadrant:
\[ \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \]
- In the third quadrant:
\[ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \]

### Step 2: Analyze \( \tan \theta = 1 \)

The tangent function is positive in the first and third quadrants. The reference angle for \( \tan \theta = 1 \) is \( \frac{\pi}{4} \). Therefore, the angles where \( \tan \theta = 1 \) can be expressed as:
- In the first quadrant:
\[ \theta = \frac{\pi}{4} \]
- In the third quadrant:
\[ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \]

### Step 3: Find the common angle

From the analysis:
- From \( \cos \theta = -\frac{1}{\sqrt{2}} \), we have possible angles \( \frac{3\pi}{4} \) and \( \frac{5\pi}{4} \).
- From \( \tan \theta = 1 \), we have possible angles \( \frac{\pi}{4} \) and \( \frac{5\pi}{4} \).

The common angle that satisfies both conditions is:
\[ \theta = \frac{5\pi}{4} \]

### Step 4: Write the general solution

The most general solution for \( \theta \) can be expressed as:
\[ \theta = \frac{5\pi}{4} + n \cdot \pi \]
where \( n \) is any integer (since the period of tangent is \( \pi \)).

### Final Answer

Thus, the most general value of \( \theta \) that satisfies both equations is:
\[ \theta = 2n\pi + \frac{5\pi}{4}, \quad n \in \mathbb{Z} \]

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