Question

$2x{\tan }^2{60}^o+3x{\sin }^2{30}^o=\frac{27{\cos }^2{45}^o}{4{\sin }^2{60}^o}$

Answer 1

Given that,

$2x{\left(tan{60}^{\circ }\right)}^2+3x{\left(sin{30}^{\circ }\right)}^2=\frac{27{\left(cos{45}^{\circ }\right)}^2}{4{\left(sin{60}^{\circ }\right)}^2}$

On substituting the values, we get

$2x{\left(\sqrt{3}\right)}^2+3x{\left(\frac{1}{2}\right)}^2=\frac{27{\left(\frac{1}{\sqrt{2}}\right)}^2}{4{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$6x+\frac{3x}{4}=\frac{9}{2}$

$2(24x+3x)=36$

$54x=36$

$x=\frac{2}{3}$

Hence, this is the answer.