Question : If $x\sin ^{2}60^{\circ}-\frac{3}{2}\sec 60^{\circ}\tan^{2}30^{\circ}+\frac{4}{5}\sin ^{2}45^{\circ}\tan ^{2}60^{\circ}=0$, then $x$ is:
Option 1: $-\frac{1}{15}$
Option 2: $–4$
Option 3: $-\frac{4}{15}$
Option 4: $–2$
Correct Answer: $-\frac{4}{15}$
Solution :
$x\sin ^{2}60^{\circ}-\frac{3}{2}\sec 60^{\circ}\tan^{2}30^{\circ}+\frac{4}{5}\sin^{2}45^{\circ}\tan ^{2}60^{\circ}=0$
$⇒x(\frac{\sqrt3}{2})^2-\frac{3}{2}×2×(\frac{1}{\sqrt3})^2+\frac{4}{5} ×\frac{1}{2}×(\sqrt{3})^2=0$
$⇒\frac{3x}{4}-1+\frac{6}{5}=0$
$⇒\frac{3x}{4}=-\frac{1}{5}$
$⇒x=-\frac{4}{15}$
Hence, the correct answer is $-\frac{4}{15}$.
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