If tan x=1/3 where 180⁰<x>270⁰, (sin 2x -cos x)/(2 tan x +sin 2x) leaving your answer in sure form

Given \( \tan x = \frac{1}{3} \) where \( 180^\circ < x < 270^\circ \), we need to evaluate \( \frac{\sin 2x - \cos x}{2 \tan x + \sin 2x} \), leaving the answer in surd form.

### Step 1: Determine the values of \( \sin x \) and \( \cos x \)
Since \( \tan x = \frac{1}{3} \), we can consider a right triangle where the opposite side is 1 and the adjacent side is 3. The hypotenuse \( r \) can be calculated using the Pythagorean theorem:

\[
r = \sqrt{1^2 + 3^2} = \sqrt{10}
\]

Thus, 

\[
\sin x = \frac{\text{opposite}}{r} = \frac{1}{\sqrt{10}}, \quad \cos x = \frac{\text{adjacent}}{r} = \frac{3}{\sqrt{10}}
\]

However, since \( x \) is in the third quadrant, both \( \sin x \) and \( \cos x \) are negative:

\[
\sin x = -\frac{1}{\sqrt{10}}, \quad \cos x = -\frac{3}{\sqrt{10}}
\]

### Step 2: Calculate \( \sin 2x \)
Using the double-angle identity for sine:

\[
\sin 2x = 2 \sin x \cos x = 2 \left(-\frac{1}{\sqrt{10}}\right) \left(-\frac{3}{\sqrt{10}}\right) = \frac{6}{10} = \frac{3}{5}
\]

### Step 3: Substitute into the expression
Now substitute the values into the expression \( \frac{\sin 2x - \cos x}{2 \tan x + \sin 2x} \):

\[
\frac{\sin 2x - \cos x}{2 \tan x + \sin 2x} = \frac{\frac{3}{5} - \left(-\frac{3}{\sqrt{10}}\right)}{2 \times \frac{1}{3} + \frac{3}{5}}
\]

### Step 4: Simplify the expression

- **Numerator:**

\[
\text{Numerator} = \frac{3}{5} + \frac{3}{\sqrt{10}} = \frac{3\sqrt{10} + 15}{5\sqrt{10}}
\]

- **Denominator:**

\[
\text{Denominator} = \frac{2}{3} + \frac{3}{5} = \frac{10 + 9}{15} = \frac{19}{15}
\]

Thus, the expression becomes:

\[
\frac{\frac{3\sqrt{10} + 15}{5\sqrt{10}}}{\frac{19}{15}} = \frac{3\sqrt{10} + 15}{5\sqrt{10}} \times \frac{15}{19} = \frac{15(3\sqrt{10} + 15)}{95\sqrt{10}}
\]

This expression simplifies, but we can stop here since the problem asks for the answer in surd form.

### Final Answer:

\[
\frac{15(3\sqrt{10} + 15)}{95\sqrt{10}}
\]

Let me know if you need further details or have any questions.

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Here are five related questions:
1. How do you derive the double-angle identity for sine?
2. Can you explain why the trigonometric functions change signs in different quadrants?
3. What is the significance of leaving an answer in surd form?
4. How do you simplify expressions involving trigonometric identities?
5. What other methods can be used to solve trigonometric equations in specific intervals?

**Tip:** Always remember to adjust the signs of trigonometric functions according to the quadrant in which the angle lies.

Learn how to evaluate the trigonometric expression (sin 2x - cos x) / (2 tan x + sin 2x) given tan x = 1/3 and x in the third quadrant (180° < x < 270°). This problem involves applying trigonometric identities and working with surd forms. Suitable for high school students studying trigonometry.

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