Solve : If `(1)/(cos theta)+(1)/(cos 3theta)=(1)/(cos 5theta)`

Explanation

To solve the equation

\[ \frac{1}{\cos \theta} + \frac{1}{\cos 3\theta} = \frac{1}{\cos 5\theta}, \]

we will follow these steps:

### Step 1: Take LCM
First, we will take the least common multiple (LCM) of the fractions on the left-hand side:

\[ \frac{\cos 3\theta + \cos \theta \cdot \cos 5\theta}{\cos \theta \cdot \cos 3\theta} = \frac{1}{\cos 5\theta}. \]

### Step 2: Cross-Multiply
Cross-multiplying gives us:

\[ \cos 3\theta + \cos \theta \cdot \cos 5\theta = \cos \theta \cdot \cos 3\theta. \]

### Step 3: Rearranging the Equation
Rearranging the equation, we have:

\[ \cos 3\theta + \cos \theta \cdot \cos 5\theta - \cos \theta \cdot \cos 3\theta = 0. \]

### Step 4: Factor Out Common Terms
We can factor out \(\cos \theta\):

\[ \cos 3\theta + \cos \theta (\cos 5\theta - \cos 3\theta) = 0. \]

### Step 5: Set Each Factor to Zero
This gives us two cases to consider:

1. \(\cos 3\theta = 0\)
2. \(\cos 5\theta - \cos 3\theta = 0\)

### Step 6: Solve \(\cos 3\theta = 0\)
For the first case, \(\cos 3\theta = 0\):

\[ 3\theta = \frac{(2n + 1)\pi}{2} \quad \text{where } n \in \mathbb{Z}. \]

Thus,

\[ \theta = \frac{(2n + 1)\pi}{6}. \]

### Step 7: Solve \(\cos 5\theta - \cos 3\theta = 0\)
For the second case, we have \(\cos 5\theta = \cos 3\theta\). This can be solved using the identity:

\[ \cos A = \cos B \implies A = 2k\pi \pm B \quad \text{where } k \in \mathbb{Z}. \]

So we have:

1. \(5\theta = 3\theta + 2k\pi\)
2. \(5\theta = -3\theta + 2k\pi\)

From the first equation:

\[ 2\theta = 2k\pi \implies \theta = k\pi. \]

From the second equation:

\[ 8\theta = 2k\pi \implies \theta = \frac{k\pi}{4}. \]

### Step 8: Combine Solutions
Combining all solutions, we have:

1. \(\theta = \frac{(2n + 1)\pi}{6}\)
2. \(\theta = k\pi\)
3. \(\theta = \frac{k\pi}{4}\)

### Final Solution
Thus, the complete solution set is:

\[ \theta = \frac{(2n + 1)\pi}{6}, \quad \theta = k\pi, \quad \theta = \frac{k\pi}{4} \quad \text{where } n, k \in \mathbb{Z}. \]