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?

Question

নিচের অপশন গুলা দেখুন

• -4
• -2
• $$ - \frac{4}{{15}}$$
• $$ - \frac{1}{{15}}$$