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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