$$ 2 \sin x-1=0 $$, find $$ x $$ if $$ 0^{\circ} \leqslant x \leqslant 360^{\circ} $$

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1. Start with the equation:  
   $$
   2 \sin x - 1 = 0
   $$

2. Add 1 to both sides:  
   $$
   2 \sin x = 1
   $$

3. Divide both sides by 2:  
   $$
   \sin x = \frac{1}{2}
   $$

4. Find all angles $$ x $$ in the interval $$ 0^\circ \leqslant x \leqslant 360^\circ $$ where $$ \sin x = \frac{1}{2} $$.  
   We know that:
   $$
   \sin 30^\circ = \frac{1}{2} \quad \text{and} \quad \sin 150^\circ = \frac{1}{2}
   $$

5. Therefore, the solutions are:  
   $$
   x = 30^\circ \quad \text{and} \quad x = 150^\circ
   $$
The solutions are $$ x = 30^\circ $$ and $$ x = 150^\circ $$.

\( 2 \sin x-1=0 \), find \( x \) if \( 0^{\circ} \leqslant x \leqslant 360^{\circ} \)

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