Đáp án:
$A=-\dfrac{8\pm 16\sqrt{3}}{11}$
Giải thích các bước giải:
$\sin^2x+\cos^2=1\\ \Rightarrow \sin^2x=1-\cos^2\\ \Leftrightarrow \sin^2x=\dfrac{3}{4}\\ \Leftrightarrow \sin x=\pm \dfrac{\sqrt{3}}{2}\\ \Rightarrow \left[\begin{array}{l} A=\dfrac{0,5+2.\dfrac{3}{4}}{0,5^2- \dfrac{\sqrt{3}}{2}}\\ A=\dfrac{0,5+2.\dfrac{3}{4}}{0,5^2- \dfrac{\sqrt{3}}{2}}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} A=\dfrac{2}{\dfrac{1-2\sqrt{3}}{4}}\\ A=\dfrac{2}{\dfrac{1+2\sqrt{3}}{4}} \end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} A=\dfrac{8}{1-2\sqrt{3}}\\ A=\dfrac{8}{1+2\sqrt{3}}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} A=-\dfrac{8+16\sqrt{3}}{11}\\ A=-\dfrac{8-16\sqrt{3}}{11}\end{array} \right.\\ \Leftrightarrow A=-\dfrac{8\pm 16\sqrt{3}}{11}$
