If sin x = sin 60^{circ} cdot cos 30^{circ} - | vedclass

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(B) Given the equation: $\sin x = \sin 60^{\circ} \cdot \cos 30^{\circ} - \cos 60^{\circ} \cdot \sin 30^{\circ}$.Using the trigonometric values: $\sin

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(B) Given the equation: $\sin x = \sin 60^{\circ} \cdot \cos 30^{\circ} - \cos 60^{\circ} \cdot \sin 30^{\circ}$.Using the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,$\cos 60^{\circ} = \frac{1}{2}$,and $\sin 30^{\circ} = \frac{1}{2}$.Substituting these values into the equation:$\sin x = \left( \frac{\sqrt{3}}{2} \right) \cdot \left( \frac{\sqrt{3}}{2} \right) - \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right)$$\sin x = \frac{3}{4} - \frac{1}{4}$$\sin x = \frac{2}{4} = \frac{1}{2}$Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\sin x = \sin 30^{\circ}$.Therefore,$x = 30^{\circ}$.

If $\sin x = \sin 60^{\circ} \cdot \cos 30^{\circ} - \cos 60^{\circ} \cdot \sin 30^{\circ}$,then $x = \ldots$ (in $^{\circ}$)

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