Cho cot x = - căn bậc hai 3 ,3pi /2 < x < 2 pi.

a) b)  Vì \(\frac{{3\pi }}{2} < x < 2\pi \) nên \(\sin x < 0;\cos x > 0\).

Ta có \(1 + {\cot ^2}x = \frac{1}{{{{\sin }^2}x}}\)\( \Rightarrow \frac{1}{{{{\sin }^2}x}} = 1 + {\left( { - \sqrt 3 } \right)^2} = 4\)\( \Rightarrow {\sin ^2}x = \frac{1}{4}\)\( \Rightarrow \sin x =  - \frac{1}{2}\).

Ta có \(\cos x = \cot x.\sin x = \left( { - \sqrt 3 } \right).\left( { - \frac{1}{2}} \right) = \frac{{\sqrt 3 }}{2}\).

c) \(\sin \left( {\frac{{4\pi }}{3} - x} \right) = \sin \frac{{4\pi }}{3}\cos x - \cos \frac{{4\pi }}{3}\sin x\)\( = \left( { - \frac{{\sqrt 3 }}{2}} \right).\frac{{\sqrt 3 }}{2} - \left( { - \frac{1}{2}} \right)\left( { - \frac{1}{2}} \right) = \frac{{ - 3}}{4} - \frac{1}{4} =  - 1\).

d) Vì \(\cot x =  - \sqrt 3 \)\( \Rightarrow \tan x =  - \frac{1}{{\sqrt 3 }}\).

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

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