Giải
Ta sẽ chứng minh: $VF = \cos{\dfrac{\pi}{24}}$
Thật vậy, ta có:
$\cos^2{\dfrac{\pi}{12}} = \dfrac{1 + \cos{\dfrac{\pi}{6}}}{2} = \dfrac{2 + \sqrt{3}}{4}$
$\Rightarrow \cos{\dfrac{\pi}{12}} = \sqrt{\dfrac{2 + \sqrt{3}}{4}} = \dfrac{\sqrt{3} + 1}{2\sqrt{2}} \,\, (\cos{\dfrac{\pi}{12}} > 0)$
Do đó:
$\cos^2{\dfrac{\pi}{24}} = \dfrac{1 + \dfrac{\sqrt{3} + 1}{2\sqrt{2}}}{2} = \dfrac{2\sqrt{2} + \sqrt{3} + 1}{4\sqrt{2}}$
$= \dfrac{(2\sqrt{2} + \sqrt{3} + 1)(2\sqrt{2} + \sqrt{3} - 1)}{4\sqrt{2}(2\sqrt{2} + \sqrt{3} - 1)} = \dfrac{(2\sqrt{2} + \sqrt{3})^2 - 1}{4\sqrt{2}(2\sqrt{2} + \sqrt{3} - 1)}$
$= \dfrac{5 + 2\sqrt{6}}{2\sqrt{2}(2\sqrt{2} + \sqrt{3} - 1)} = \dfrac{1}{2\sqrt{2}(2\sqrt{2} + \sqrt{3} - 1)(5 - 2\sqrt{6})}$
$\Rightarrow \cos{\dfrac{\pi}{24}} = \dfrac{1}{\sqrt{2\sqrt{2}(2\sqrt{2} + \sqrt{3} - 1)(5 - 2\sqrt{6})}} \,\, (\cos{\dfrac{\pi}{24}} > 0)$
Ta thấy:
$2\sqrt{2}(2\sqrt{2} + \sqrt{3} - 1)(5 - 2\sqrt{6}) = (8 + 2\sqrt{6} - 2\sqrt{2})(5 - 2\sqrt{6})$
$= \left[(5 + 2\sqrt{6}) + (3 - 2\sqrt{2})\right](5 - 2\sqrt{6})$
$= 1 + (\sqrt{2} - 1)^2(\sqrt{3} - \sqrt{2})^2$
$= 1 + \left(\sqrt{6} - 2 - \sqrt{3} + \sqrt{2}\right)^2$
Vậy:
$VF = \dfrac{1}{\sqrt{1 + \left(\sqrt{6} - 2 - \sqrt{3} + \sqrt{2}\right)^2}} = \cos{\dfrac{\pi}{24}}$
Khi đó, phương trình có nghiệm:
$x = \pm \dfrac{\pi}{24} + 2k\pi \,\, (k \in Z)$
P/S: (Có dùng máy tính để nhẩm nghiệm)
Bài viết đã được chỉnh sửa nội dung bởi Ispectorgadget: 30-08-2012 - 20:41