## Trigonometry 7th Edition

- $\frac{3\sqrt 3}{2}$ or $- \frac{3}{2} \sqrt 3$
Given expression = -3 $\sin2x$ = -3 $\sin2 (30^{\circ})$ ( replacing $x$ with $30^{\circ}$) = -3 $\sin60^{\circ}$ = -3 $\times (\frac{\sqrt 3}{2})$ ( substituting exact value of $\sin 60^{\circ}$) = - $\frac{3\sqrt 3}{2}$ or $- \frac{3}{2} \sqrt 3$