Phương pháp giải:

\(\dfrac{1}{{{{\cos }^2}x}} = {\tan ^2}x + 1\)

\(\dfrac{\pi }{2} < x < \pi \) thì \(\cos x < 0,\sin x > 0\)

\(\cos 2x = 2{\cos ^2}x - 1\)

\(\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a.\tan b}}\)

\(\sin \left( {a + b} \right)\)\( = \sin a.\cos b + \cos a.\sin b\)

Lời giải chi tiết:

Ta có \(\dfrac{1}{{{{\cos }^2}x}} = {\tan ^2}x + 1 = \dfrac{{25}}{{16}}\)

\( =  > {\cos ^2}x = \dfrac{{16}}{{25}},{\sin ^2}x = \dfrac{9}{{25}}\)

Do \(\dfrac{\pi }{2} < x < \pi \) nên \(\cos x < 0,\sin x > 0\)

=> \(\cos x = \dfrac{{ - 4}}{5},\sin x = \dfrac{3}{5}\)

\(\begin{array}{l}\cos 2x = 2{\cos ^2}x - 1 = \dfrac{7}{{25}}\\\tan \left( {x - \dfrac{\pi }{4}} \right) = \dfrac{{\tan x - \tan \dfrac{\pi }{4}}}{{1 + \tan x.\tan \dfrac{\pi }{4}}}\\ =  - 7\\\sin \left( {\dfrac{\pi }{3} + x} \right)\\ = \sin \dfrac{\pi }{3}.\cos x + \cos \dfrac{\pi }{3}.\sin x\\ = \dfrac{{\sqrt 3 }}{2}.\left( { - \dfrac{4}{5}} \right) + \dfrac{1}{2}.\dfrac{3}{5} = \dfrac{{3 - 4\sqrt 3 }}{{10}}\end{array}\)