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?

• A. $ - \frac{1}{{15}}$$
• B. 4
• C. $ - \frac{4}{{15}}$$
• D. 2