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

Explanation

To solve the equation \(3(\sec^2 \theta + \tan^2 \theta) = 5\), we can use the trigonometric identity that relates secant and tangent:

\[ \sec^2 \theta = 1 + \tan^2 \theta \]

### Step 1: Substitute the identity
Substituting the identity into the equation gives us:

\[ 3(1 + \tan^2 \theta + \tan^2 \theta) = 5 \]

This simplifies to:

\[ 3(1 + 2\tan^2 \theta) = 5 \]

### Step 2: Expand and simplify
Now, expand the left side:

\[ 3 + 6\tan^2 \theta = 5 \]

### Step 3: Isolate the term with \(\tan^2 \theta\)
Next, isolate the \(6\tan^2 \theta\) term:

\[ 6\tan^2 \theta = 5 - 3 \]
\[ 6\tan^2 \theta = 2 \]

### Step 4: Solve for \(\tan^2 \theta\)
Now, divide both sides by 6:

\[ \tan^2 \theta = \frac{2}{6} = \frac{1}{3} \]

### Step 5: Take the square root
Taking the square root of both sides gives:

\[ \tan \theta = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3} \]

### Step 6: Find the general solutions for \(\theta\)
The general solutions for \(\tan \theta = \frac{\sqrt{3}}{3}\) and \(\tan \theta = -\frac{\sqrt{3}}{3}\) are:

\[ \theta = n\pi + \frac{\pi}{6} \quad \text{and} \quad \theta = n\pi - \frac{\pi}{6} \quad (n \in \mathbb{Z}) \]

Thus, the complete solution set is:

\[ \theta = n\pi + \frac{\pi}{6}, \quad n \in \mathbb{Z} \quad \text{and} \quad \theta = n\pi - \frac{\pi}{6}, \quad n \in \mathbb{Z} \]