a) Tìm tất cả các nghiệm nguyên của phương trình sau:
        $\cos \left[ {\frac{\pi }{8}\left( {3x - \sqrt {9{x^2} + 160x + 800} } \right)} \right] = 1$
b) Biết rằng $\sin 15^0= \frac{\sqrt{6 } -\sqrt{2 } }{4}$. Tính các tỉ số lượng giác của góc $15^0$.
c) Giải phương trình: \(\sin 4x - \cos 4x = 1 + 4\sqrt 2 \sin \left( {x - \frac{\pi }{4}} \right)\)
d) $\sin^2 x(\tan x+1)=3\sin x(\cos x-\sin x)+3$
e) Giải phương trình lượng giác: $2\sqrt{2}\sin(x+\frac{\pi}{4} )=\frac{1}{\sin x}+\frac{1}{\cos x} $
 
b. Ta có: $\cos^215^0=1-\sin^215^0=1-\left (\frac{\sqrt{ 6}-\sqrt{ 2}  }{4}\right)^2=\frac{8+4\sqrt{3 } }{16}=\frac{(\sqrt{3 } +1)^2}{8}   $
Từ đó $\cos 15^0=\sqrt{ \frac{(\sqrt{ 3}+1)^2 }{8} } =\frac{\sqrt{ 3}+1 }{2\sqrt{ 2} }=\frac{(\sqrt{3 }+1)\sqrt{2 }}{4}=\frac{\sqrt{6 }+\sqrt{ 2}  }{4} $
$\tan15^0=\frac{\sin15^0}{\cos 15^0} =\frac{\sqrt{6 }-\sqrt{2 }  }{\sqrt{6 }+\sqrt{2 }  } =\frac{(\sqrt{6 }-\sqrt{ 2})^2  }{4}=2-\sqrt{ 3} $
$\cot 15^0=\frac{1}{2-\sqrt{3 } }=2+\sqrt{3 } $

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