Question : If $4 \sin ^2 \mathrm{A}-3=0$ and $0 \leq \mathrm{A} \leq 90^{\circ}$, then $3 \sin \mathrm{A}-4 \sin ^3 \mathrm{A}$ is:
Option 1: $\frac{1}{2}$
Option 2: $\sqrt{\frac{3}{2}}$
Option 3: $0$
Option 4: $1$
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Correct Answer: $0$
Solution : $4 \sin ^2 \mathrm{A}-3=0$ $⇒\sin ^2 A = \frac{3}{4}$ $⇒\sin A = \frac{\sqrt{3}}{2}$ $3 \sin \mathrm{A}-4 \sin ^3 \mathrm{A}$ $=3 (\frac{\sqrt{3}}{2}) - 4 (\frac{\sqrt{3}}{2})^3 $ $=3 (\frac{\sqrt{3}}{2}) - 4 (\frac{\sqrt{3}}{2})^3 $ $=3 (\frac{\sqrt{3}}{2}) - 4 (\frac{3\sqrt{3}}{8}) $ $=(\frac{3\sqrt{3}}{2}) - (\frac{3\sqrt{3}}{2}) $ $= 0$ Hence, the correct answer is $0$.
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