Find the general solution of :
`sqrt(3)cosx-sinx=1`

Answer

To find the general solution of the equation \( \sqrt{3} \cos x - \sin x = 1 \), we can follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ \sqrt{3} \cos x - \sin x = 1 \]
We can rearrange it as:
\[ \sqrt{3} \cos x - \sin x - 1 = 0 \]

### Step 2: Divide by 2
Next, we divide the entire equation by 2:
\[ \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = \frac{1}{2} \]

### Step 3: Recognize trigonometric identities
We know that:
\[ \frac{\sqrt{3}}{2} = \sin \frac{\pi}{3} \quad \text{and} \quad \frac{1}{2} = \cos \frac{\pi}{3} \]
Thus, we can rewrite the equation as:
\[ \sin \frac{\pi}{3} \cos x - \cos \frac{\pi}{3} \sin x = \frac{1}{2} \]

### Step 4: Use the sine subtraction formula
Using the sine subtraction formula:
\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]
we can express the left side as:
\[ \sin\left(\frac{\pi}{3} - x\right) = \frac{1}{2} \]

### Step 5: Solve for \( x \)
Now, we need to find the angles for which the sine is \( \frac{1}{2} \):
\[ \frac{\pi}{3} - x = n\pi + \frac{\pi}{6} \quad \text{for } n \in \mathbb{Z} \]
This gives us two cases:
1. \( \frac{\pi}{3} - x = n\pi + \frac{\pi}{6} \)
2. \( \frac{\pi}{3} - x = n\pi + \frac{5\pi}{6} \)

### Step 6: Solve each case for \( x \)

**Case 1:**
\[ \frac{\pi}{3} - x = n\pi + \frac{\pi}{6} \]
Rearranging gives:
\[ x = \frac{\pi}{3} - n\pi - \frac{\pi}{6} \]
Combining the fractions:
\[ x = \frac{2\pi}{6} - \frac{\pi}{6} - n\pi = \frac{\pi}{6} - n\pi \]

**Case 2:**
\[ \frac{\pi}{3} - x = n\pi + \frac{5\pi}{6} \]
Rearranging gives:
\[ x = \frac{\pi}{3} - n\pi - \frac{5\pi}{6} \]
Combining the fractions:
\[ x = \frac{2\pi}{6} - \frac{5\pi}{6} - n\pi = -\frac{3\pi}{6} - n\pi = -\frac{\pi}{2} - n\pi \]

### Final General Solution
Thus, the general solutions are:
\[ x = \frac{\pi}{6} - n\pi \quad \text{and} \quad x = -\frac{\pi}{2} - n\pi \quad \text{for } n \in \mathbb{Z} \]