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The problem presented is:

Given that:
\[
\sqrt{3} \tan \theta = 3 \sin \theta
\]
You are asked to find:
\[
\sin 2\theta - \cos 2\theta
\]

### Step 1: Simplifying the given equation
Start with:
\[
\sqrt{3} \tan \theta = 3 \sin \theta
\]
Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), so substitute this into the equation:
\[
\sqrt{3} \frac{\sin \theta}{\cos \theta} = 3 \sin \theta
\]

### Step 2: Solve for \(\cos \theta\)
Now, divide both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):
\[
\frac{\sqrt{3}}{\cos \theta} = 3
\]
Multiply both sides by \(\cos \theta\) and solve for \(\cos \theta\):
\[
\sqrt{3} = 3 \cos \theta
\]
\[
\cos \theta = \frac{\sqrt{3}}{3}
\]

### Step 3: Calculate \(\sin \theta\)
Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), substitute \(\cos \theta = \frac{\sqrt{3}}{3}\):
\[
\sin^2 \theta + \left(\frac{\sqrt{3}}{3}\right)^2 = 1
\]
\[
\sin^2 \theta + \frac{3}{9} = 1
\]
\[
\sin^2 \theta = 1 - \frac{1}{3}
\]
\[
\sin^2 \theta = \frac{2}{3}
\]
\[
\sin \theta = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{6}}{3}
\]

### Step 4: Calculate \(\sin 2\theta - \cos 2\theta\)
Now, use the double angle formulas:
- \(\sin 2\theta = 2 \sin \theta \cos \theta\)
- \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\)

First, calculate \(\sin 2\theta\):
\[
\sin 2\theta = 2 \times \frac{\sqrt{6}}{3} \times \frac{\sqrt{3}}{3} = \frac{2\sqrt{18}}{9} = \frac{6}{9} = \frac{2}{3}
\]

Now, calculate \(\cos 2\theta\):
\[
\cos 2\theta = \cos^2 \theta - \sin^2 \theta = \left(\frac{\sqrt{3}}{3}\right)^2 - \left(\frac{\sqrt{6}}{3}\right)^2 = \frac{3}{9} - \frac{6}{9} = -\frac{1}{3}
\]

Finally, \(\sin 2\theta - \cos 2\theta\):
\[
\sin 2\theta - \cos 2\theta = \frac{2}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3} + \frac{1}{3} = 1
\]

### Final Answer:
The value of \(\sin 2\theta - \cos 2\theta\) is \(1\).

Would you like further details or clarification on any of the steps? 

Here are 5 related questions for further exploration:
1. What is the value of \(\tan 2\theta\) given \(\cos \theta = \frac{\sqrt{3}}{3}\)?
2. How does the double angle formula for cosine differ from that for sine?
3. What happens if \(\theta\) is in a different quadrant? How would the signs of \(\sin \theta\) and \(\cos \theta\) change?
4. Can you derive the half-angle identities from the double-angle identities?
5. How do trigonometric identities apply in solving non-right triangles?

**Tip:** Always remember to check which quadrant your angle is in when solving trigonometric equations, as it affects the signs of sine and cosine.

If √3 tan θ = 3 sin θ, then sin 2θ - cos 2θ = ?

Learn how to solve the trigonometric equation √3 tan θ = 3 sin θ and calculate sin 2θ - cos 2θ. This problem covers key trigonometric concepts including the Pythagorean identity and double angle formulas. Suitable for students in Grades 11-12.

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