2 sin x cos x =
√
3 − 2sqrt3 sin2
x
2sqrt3 sin2
x + sin 2x =
√
3
√
3
3
sin 2x = 1 − 2 sin2
x
√
3
3
sin 2x = cos 2x
sin 2x =
3
√
3
cos 2x | : cos 2x
tan 2x =
3
√
3
2x = 60◦ + n · 180◦
x = 30◦ + n · 90◦

The problem you've provided involves solving a trigonometric equation that follows from an identity related to the sine and cosine functions, eventually leading to an equation for \( x \).

### Breakdown of Steps:

1. **Step 1: Begin with the identity**:
   \[
   2 \sin x \cos x = \sin 2x
   \]
   This is a standard trigonometric identity, stating that \( \sin 2x = 2 \sin x \cos x \).

2. **Step 2: Manipulate the given equation**:
   \[
   2 \sqrt{3} \sin^2 x + \sin 2x = \sqrt{3}
   \]
   We substitute \( \sin 2x = 2 \sin x \cos x \) into this equation:
   \[
   2 \sqrt{3} \sin^2 x + 2 \sin x \cos x = \sqrt{3}
   \]

3. **Step 3: Express \( \sin 2x \) in terms of \( \cos 2x \)**:
   The equation \( \sin 2x = \cos 2x \) implies:
   \[
   \tan 2x = 1
   \]
   Solving \( \tan 2x = 1 \) gives:
   \[
   2x = 45^\circ + n \cdot 180^\circ
   \]
   Dividing both sides by 2:
   \[
   x = 22.5^\circ + n \cdot 90^\circ
   \]

4. **Step 4: Solve for general solutions of \( x \)**:
   In another step, you have \( 2x = 60^\circ + n \cdot 180^\circ \), leading to:
   \[
   x = 30^\circ + n \cdot 90^\circ
   \]

### Summary:
- The solution leads to the general form \( x = 30^\circ + n \cdot 90^\circ \), where \( n \) is an integer.

Would you like further details or a breakdown of any specific steps?

Here are 5 related questions to further explore this topic:
1. How do you prove the identity \( \sin 2x = 2 \sin x \cos x \)?
2. What are the general solutions for trigonometric equations involving \( \tan x = 1 \)?
3. How do you convert between degrees and radians in trigonometric equations?
4. How can you derive other double-angle identities, like \( \cos 2x \)?
5. What is the general method for solving trigonometric equations involving multiple angles?

**Tip:** When solving trigonometric equations, always consider both the principal solution and general solutions based on the periodicity of the trigonometric functions.

2 sin x cos x = √3 − 2√3 sin²x
2√3 sin²x + sin 2x = √3
√3
3 sin 2x = 1 − 2 sin²x
√3
3 sin 2x = cos 2x
sin 2x = 3
√3 cos 2x
tan 2x = √3
2x = 60° + n · 180°
x = 30° + n · 90°

Explore how to solve a trigonometric equation using double angle identities. Learn the steps to manipulate the equation and find the general solution x = 30° + n · 90°. Ideal for high school students studying advanced trigonometry concepts.

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