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