If x ^260^ - 3/2 sec60^ . ta n^230^ + 4/5 ^245^ . ta n^260^ = 0, then x is?

Answer: Option C

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