Solve: `cottheta` `+cos e ctheta=sqrt(3)`

Explanation

To solve the equation \( \cot \theta + \csc \theta = \sqrt{3} \), we will follow these steps:

### Step 1: Rewrite the trigonometric functions
We know that:
\[ \cot \theta = \frac{\cos \theta}{\sin \theta} \quad \text{and} \quad \csc \theta = \frac{1}{\sin \theta} \]
Substituting these into the equation gives:
\[ \frac{\cos \theta}{\sin \theta} + \frac{1}{\sin \theta} = \sqrt{3} \]

### Step 2: Combine the fractions
Since both terms on the left have the same denominator, we can combine them:
\[ \frac{\cos \theta + 1}{\sin \theta} = \sqrt{3} \]

### Step 3: Cross-multiply
Cross-multiplying gives:
\[ \cos \theta + 1 = \sqrt{3} \sin \theta \]

### Step 4: Square both sides
To eliminate the square root, we square both sides:
\[ (\cos \theta + 1)^2 = 3 \sin^2 \theta \]

### Step 5: Expand both sides
Expanding the left side:
\[ \cos^2 \theta + 2 \cos \theta + 1 = 3 \sin^2 \theta \]
Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we substitute for \( \sin^2 \theta \):
\[ \cos^2 \theta + 2 \cos \theta + 1 = 3(1 - \cos^2 \theta) \]

### Step 6: Rearrange the equation
Rearranging gives:
\[ \cos^2 \theta + 2 \cos \theta + 1 = 3 - 3 \cos^2 \theta \]
Combining like terms results in:
\[ 4 \cos^2 \theta + 2 \cos \theta - 2 = 0 \]

### Step 7: Simplify the equation
Dividing the entire equation by 2:
\[ 2 \cos^2 \theta + \cos \theta - 1 = 0 \]

### Step 8: Factor the quadratic equation
We can factor this quadratic:
\[ (2 \cos \theta + 1)(\cos \theta - 1) = 0 \]

### Step 9: Solve for \( \cos \theta \)
Setting each factor to zero gives us two cases:
1. \( 2 \cos \theta + 1 = 0 \) which leads to \( \cos \theta = -\frac{1}{2} \)
2. \( \cos \theta - 1 = 0 \) which leads to \( \cos \theta = 1 \)

### Step 10: Find the angles
For \( \cos \theta = -\frac{1}{2} \):
\[ \theta = \frac{2\pi}{3} + 2n\pi \quad \text{and} \quad \theta = \frac{4\pi}{3} + 2n\pi \]
For \( \cos \theta = 1 \):
\[ \theta = 2n\pi \]

### Final Solution
Thus, the general solutions for \( \theta \) are:
\[ \theta = 2n\pi, \quad \theta = \frac{2\pi}{3} + 2n\pi, \quad \theta = \frac{4\pi}{3} + 2n\pi \]

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