Solve `3(sec^(2) theta+tan^(2) theta)=5`.

Explanation

To solve the equation \( 3(\sec^2 \theta + \tan^2 \theta) = 5 \), we can follow these steps:

### Step 1: Use the identity for secant and tangent
We know the trigonometric identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta \]
Substituting this identity into the equation gives:
\[ 3((1 + \tan^2 \theta) + \tan^2 \theta) = 5 \]

### Step 2: Simplify the equation
This simplifies to:
\[ 3(1 + 2\tan^2 \theta) = 5 \]
Expanding this, we have:
\[ 3 + 6\tan^2 \theta = 5 \]

### Step 3: Rearrange the equation
Now, we can rearrange the equation to isolate the term with \(\tan^2 \theta\):
\[ 6\tan^2 \theta = 5 - 3 \]
\[ 6\tan^2 \theta = 2 \]

### Step 4: Solve for \(\tan^2 \theta\)
Dividing both sides by 6 gives:
\[ \tan^2 \theta = \frac{2}{6} = \frac{1}{3} \]

### Step 5: Take the square root
Now, taking the square root of both sides, we find:
\[ \tan \theta = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} \]

### Step 6: Find the general solution for \(\theta\)
We know that:
\[ \tan \theta = \frac{1}{\sqrt{3}} \implies \theta = \frac{\pi}{6} + n\pi \quad \text{(for the positive case)} \]
\[ \tan \theta = -\frac{1}{\sqrt{3}} \implies \theta = \frac{5\pi}{6} + n\pi \quad \text{(for the negative case)} \]
where \( n \) is any integer.

### Final Solution
Thus, the complete solution for \(\theta\) is:
\[ \theta = \frac{\pi}{6} + n\pi \quad \text{and} \quad \theta = \frac{5\pi}{6} + n\pi \]