Cho \(\cos 2x = \frac{1}{7},\pi < x < \frac{{3\pi }}{2}\). Khi đó

a) \(\cos x + \cos 3x = \frac{{4\sqrt 7 }}{{49}}\).

b) \(\cos x = - \frac{{2\sqrt 7 }}{7}\).

c) \(\sin x = \frac{{\sqrt {21} }}{7}\).

d) \(\tan \left( {x - \frac{\pi }{4}} \right) = - 7 + 4\sqrt 3 \).

Ta có \({\cos ^2}x = \frac{{1 + \cos 2x}}{2} = \frac{{1 + \frac{1}{7}}}{2} = \frac{4}{7}\).

Vì \(\pi  < x < \frac{{3\pi }}{2}\) nên \(\cos x < 0 \Rightarrow \cos x =  - \frac{{2\sqrt 7 }}{7}\).

a) \(\cos x + \cos 3x = 2\cos 2x\cos x = 2.\frac{1}{7}.\frac{{ - 2\sqrt 7 }}{7} = \frac{{ - 4\sqrt 7 }}{{49}}\).

b) \(\cos x =  - \frac{{2\sqrt 7 }}{7}\).

c) \({\sin ^2}x = 1 - {\cos ^2}x = 1 - \frac{4}{7} = \frac{3}{7}\) mà sinx < 0 nên \(\sin x =  - \frac{{\sqrt {21} }}{7}\).

d) Có \(\tan x = \frac{{\sin x}}{{\cos x}} = \frac{{ - \sqrt {21} }}{7}:\frac{{ - 2\sqrt 7 }}{7} = \frac{{\sqrt 3 }}{2}\).

\(\tan \left( {x - \frac{\pi }{4}} \right) = \frac{{\tan x - \tan \frac{\pi }{4}}}{{1 + \tan x\tan \frac{\pi }{4}}} = \frac{{\frac{{\sqrt 3 }}{2} - 1}}{{1 + \frac{{\sqrt 3 }}{2}}} =  - 7 + 4\sqrt 3 \).

Đáp án: a) Sai;   b) Đúng;   c) Sai;   d) Đúng.