If $$x{\sin ^2}{60^ \circ }$$  - $$\frac{3}{2}{\text{sec}}{60^ \circ }$$ . $${\text{ta}}{{\text{n}}^2}{30^ \circ }$$  + $$\frac{4}{5}{\sin ^2}{45^ \circ }$$ . $${\text{ta}}{{\text{n}}^2}{60^ \circ }$$  = 0, then x is?

Solution(By Examveda Team)

$${\text{ }}x{\sin ^2}{60^ \circ } - \frac{3}{2}{\text{sec}}{60^ \circ }{\text{.ta}}{{\text{n}}^2}{30^ \circ } + $$       $$\frac{4}{5}{\sin ^2}{45^ \circ }.$$   $${\text{ta}}{{\text{n}}^2}{60^ \circ }$$   $$ = 0$$
$$ \Rightarrow {\text{ }}x{\left( {\frac{{\sqrt 3 }}{2}} \right)^2} - \frac{3}{2} \times {\text{2}} \times {\left( {\frac{1}{{\sqrt 3 }}} \right)^2}{\text{ + }}$$       $$\frac{4}{5}{\left( {\frac{1}{{\sqrt 2 }}} \right)^2} \times $$   $${\left( {\sqrt 3 } \right)^2} = 0$$
$$\eqalign{ & \Rightarrow \frac{{3x}}{4} - \frac{3}{2} \times 2 \times \frac{1}{3} + \frac{4}{5} \times \frac{1}{2} \times 3 = 0 \cr & \Rightarrow \frac{{3x}}{4} - 1 + \frac{6}{5} = 0 \cr & \Rightarrow \frac{{3x}}{4} = 1 - \frac{6}{5} \cr & \Rightarrow \frac{{5 - 6}}{5} \cr & \Rightarrow \frac{{ - 1}}{5} \cr & \therefore x = - \frac{1}{5} \times \frac{4}{3} \cr & \,\,\,\,\,\,\,\,\,\, = - \frac{4}{{15}} \cr} $$