Find the general solution of :
`cosx+sinx=1`

Answer

To find the general solution of the equation \( \cos x + \sin x = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos x + \sin x = 1 \]

### Step 2: Divide by \( \sqrt{2} \)
To simplify the equation, we divide each term by \( \sqrt{2} \):
\[ \frac{\cos x}{\sqrt{2}} + \frac{\sin x}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]
This can be rewritten as:
\[ \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} \]

### Step 3: Recognize the cosine form
Notice that \( \frac{1}{\sqrt{2}} \) is equal to \( \cos \frac{\pi}{4} \) and \( \sin \frac{\pi}{4} \). Thus, we can express the left-hand side in terms of cosine:
\[ \cos \frac{\pi}{4} \cos x + \sin \frac{\pi}{4} \sin x = \frac{1}{\sqrt{2}} \]
This can be rewritten using the cosine of a difference:
\[ \cos\left(x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

### Step 4: Solve for \( x - \frac{\pi}{4} \)
The general solution for \( \cos \theta = k \) is given by:
\[ \theta = 2n\pi \pm \cos^{-1}(k) \]
In our case, \( k = \frac{1}{\sqrt{2}} \), so:
\[ x - \frac{\pi}{4} = 2n\pi \pm \frac{\pi}{4} \]

### Step 5: Solve for \( x \)
Now we can solve for \( x \):
1. For the positive case:
\[ x - \frac{\pi}{4} = 2n\pi + \frac{\pi}{4} \]
\[ x = 2n\pi + \frac{\pi}{2} \]

2. For the negative case:
\[ x - \frac{\pi}{4} = 2n\pi - \frac{\pi}{4} \]
\[ x = 2n\pi \]

### Final General Solution
Combining both cases, we have the general solutions:
\[ x = 2n\pi + \frac{\pi}{2} \quad \text{and} \quad x = 2n\pi \]
where \( n \) is any integer.