ham so luong giac kho!

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kieuoanh_1510

[tex]2\sqrt{3}.sin\left ( x-\frac{\Pi }{8} \right ).cos\left ( x-\frac{ \Pi }{8}) +2.cos^2\right (x-\frac{\Pi }{8})=\sqrt{3}+1\\ \Leftrightarrow \sqrt{3}sin\left ( 2x-\frac{\Pi }{4} \right )+1+cos\left ( 2x-\frac{\Pi }{4} \right )=\sqrt{3}+1\\\Leftrightarrow \frac{\sqrt{3}}{2}sin\left ( 2x-\frac{\Pi }{4} \right )+\frac{1}{2}cos\left ( 2x-\Pi /4 \right )=\frac{\sqrt{3}}{2}\\\Leftrightarrow sin\frac({\Pi }/{3})sin\left ( 2x-\frac{\Pi }{4} \right ) + cos\frac({\Pi }/{3}).cos\left ( 2x-\Pi /4 \right )=\frac{\sqrt{3}}{2} --->dang co ban[/tex]
 
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