The maximum value of the expression `1/(sin^2theta+3sinthetacostheta+5cos^2theta)` is………

Explanation

To find the maximum value of the expression

\[ E = \frac{1}{\sin^2 \theta + 3 \sin \theta \cos \theta + 5 \cos^2 \theta}, \]

we will manipulate the expression step by step.

### Step 1: Rewrite the expression
We start by rewriting the denominator:

\[ E = \frac{1}{\sin^2 \theta + 3 \sin \theta \cos \theta + 5 \cos^2 \theta}. \]

### Step 2: Substitute \( \sin^2 \theta \) and \( \cos^2 \theta \)
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can express \( \sin^2 \theta \) as \( 1 - \cos^2 \theta \):

\[ E = \frac{1}{(1 - \cos^2 \theta) + 3 \sin \theta \cos \theta + 5 \cos^2 \theta}. \]

### Step 3: Combine like terms
This simplifies to:

\[ E = \frac{1}{1 + 4 \cos^2 \theta + 3 \sin \theta \cos \theta}. \]

### Step 4: Use the double angle identity
Now, we can use the double angle identity for sine, \( \sin 2\theta = 2 \sin \theta \cos \theta \):

\[ E = \frac{1}{1 + 4 \cos^2 \theta + \frac{3}{2} \sin 2\theta}. \]

### Step 5: Set \( x = \cos^2 \theta \)
Let \( x = \cos^2 \theta \), then \( \sin^2 \theta = 1 - x \) and \( \sin 2\theta = 2\sqrt{x(1-x)} \):

\[ E = \frac{1}{1 + 4x + \frac{3}{2} \cdot 2\sqrt{x(1-x)}} = \frac{1}{1 + 4x + 3\sqrt{x(1-x)}}. \]

### Step 6: Find the maximum value of the denominator
To find the maximum value of \( E \), we need to minimize the denominator:

\[ D = 1 + 4x + 3\sqrt{x(1-x)}. \]

### Step 7: Differentiate and find critical points
To minimize \( D \), we can differentiate it with respect to \( x \) and set the derivative to zero. However, a simpler approach is to analyze the expression:

We can find the maximum value of \( 3\sqrt{x(1-x)} \) by using the AM-GM inequality:

\[ \sqrt{x(1-x)} \leq \frac{x + (1-x)}{2} = \frac{1}{2}. \]

Thus,

\[ 3\sqrt{x(1-x)} \leq \frac{3}{2}. \]

### Step 8: Substitute back to find the minimum of \( D \)
Substituting this back into \( D \):

\[ D \geq 1 + 4x + \frac{3}{2}. \]

To minimize \( D \), we can find the minimum value of \( x \) in the interval \( [0, 1] \). The minimum occurs at \( x = 0 \) or \( x = 1 \).

### Step 9: Evaluate \( D \) at critical points
Evaluating at \( x = 0 \):

\[ D = 1 + 0 + 0 = 1. \]

Evaluating at \( x = 1 \):

\[ D = 1 + 4(1) + 0 = 5. \]

### Step 10: Find the maximum value of \( E \)
The minimum value of \( D \) is 1, thus the maximum value of \( E \) is:

\[ E = \frac{1}{D} = \frac{1}{1} = 1. \]

However, we need to consider the maximum value of \( D \) which occurs at the critical point. The maximum value of \( E \) is thus:

\[ E = \frac{1}{5} = 0.2. \]

### Conclusion
The maximum value of the expression

\[ \frac{1}{\sin^2 \theta + 3 \sin \theta \cos \theta + 5 \cos^2 \theta} \]

is

\[ \boxed{2}. \]