Solve ` cos x - sqrt3 sin x =1, 0^(@) le x le 360 ^(@).`

Answer

To solve the equation \( \cos x - \sqrt{3} \sin x = 1 \) for \( 0^\circ \leq x \leq 360^\circ \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \cos x - \sqrt{3} \sin x = 1 \]

### Step 2: Divide by 2
To simplify the equation, we can divide both sides by 2:
\[ \frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x = \frac{1}{2} \]

### Step 3: Recognize trigonometric identities
We can express \( \frac{1}{2} \) and \( \frac{\sqrt{3}}{2} \) in terms of sine and cosine:
\[ \frac{1}{2} = \sin \frac{\pi}{6} \quad \text{and} \quad \frac{\sqrt{3}}{2} = \cos \frac{\pi}{6} \]
Thus, we can rewrite the equation as:
\[ \sin \frac{\pi}{6} \cos x - \cos \frac{\pi}{6} \sin x = \sin \frac{\pi}{6} \]

### 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 rewrite the left-hand side:
\[ \sin\left(\frac{\pi}{6} - x\right) = \sin \frac{\pi}{6} \]

### Step 5: Set up the general solution
The general solution for \( \sin A = \sin B \) is given by:
\[ A = n\pi + (-1)^n B \]
Applying this to our equation:
\[ \frac{\pi}{6} - x = n\pi + (-1)^n \frac{\pi}{6} \]

### Step 6: Solve for \( x \)
Now we can solve for \( x \):
1. For \( n = 0 \):
\[ \frac{\pi}{6} - x = 0 + \frac{\pi}{6} \implies x = 0 \]

2. For \( n = 1 \):
\[ \frac{\pi}{6} - x = \pi - \frac{\pi}{6} \implies \frac{\pi}{6} - x = \frac{5\pi}{6} \implies -x = \frac{5\pi}{6} - \frac{\pi}{6} \implies -x = \frac{4\pi}{6} \implies x = -\frac{4\pi}{6} = -\frac{2\pi}{3} \]
Since \( -\frac{2\pi}{3} \) is not in the range \( 0^\circ \) to \( 360^\circ \), we convert it:
\[ x = 360^\circ - \frac{2\pi}{3} \times \frac{180^\circ}{\pi} = 360^\circ - 120^\circ = 240^\circ \]

### Step 7: Final solutions
Thus, the solutions to the equation \( \cos x - \sqrt{3} \sin x = 1 \) in the interval \( 0^\circ \leq x \leq 360^\circ \) are:
\[ x = 0^\circ \quad \text{and} \quad x = 240^\circ \]