re's a precalculus question for you: Solve the equation $$ 2 \sin (x) \cos (x)= $$ $$ s(x) $$ for all values of $$ x $$ in the interval $$ [0,2 \pi) $$.

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1. Write the given equation:
   $$
   2 \sin(x)\cos(x)= s(x)
   $$
   Recognize that the notation $$ s(x) $$ represents $$\sin(x)$$, so the equation becomes:
   $$
   2 \sin(x)\cos(x)= \sin(x)
   $$

2. Use the double-angle identity:
   $$
   \sin(2x)=2\sin(x)\cos(x)
   $$
   Thus, the equation can be rewritten as:
   $$
   \sin(2x)=\sin(x)
   $$

3. Rearrange the equation:
   $$
   \sin(2x)-\sin(x)=0
   $$
   Factor out $$\sin(x)$$ using the expression for $$\sin(2x)$$:
   $$
   2\sin(x)\cos(x)-\sin(x)=0
   $$
   Factor $$\sin(x)$$ from the equation:
   $$
   \sin(x)(2\cos(x)-1)=0
   $$

4. Set each factor equal to zero:
   
   **Case 1:**
   $$
   \sin(x)=0
   $$
   The solutions in the interval $$[0,2\pi)$$ are:
   $$
   x=0 \quad \text{and} \quad x=\pi
   $$
   
   **Case 2:**
   $$
   2\cos(x)-1=0 \quad \Rightarrow \quad \cos(x)=\frac{1}{2}
   $$
   The solutions in the interval $$[0,2\pi)$$ are:
   $$
   x=\frac{\pi}{3} \quad \text{and} \quad x=\frac{5\pi}{3}
   $$

5. The complete solution set is:
   $$
   \boxed{0,\ \frac{\pi}{3},\ \pi,\ \frac{5\pi}{3}}
   $$
The solutions are $$ x = 0, \frac{\pi}{3}, \pi, \frac{5\pi}{3} $$.

re's a precalculus question for you: Solve the equation \( 2 \sin (x) \cos (x)= \) \( s(x) \) for all values of \( x \) in the interval \( [0,2 \pi) \).

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