Формула произведения косинусов:

[tex]\cos\alpha \cos\beta =\dfrac{1}{2} \left(\cos(\alpha +\beta )+\cos(\alpha -\beta )\right)[/tex]

Формула суммы косинусов:

[tex]\cos\alpha+ \cos\beta =2\cos\dfrac{\alpha +\beta }{2} \cos\dfrac{\alpha-\beta }{2}[/tex]

Преобразуем:

[tex]\cos12x-\cos6x-2\cos7x\cos5x=[/tex]

[tex]=\cos12x-\cos6x-2\cdot\dfrac{1}{2}\left( \cos(7x+5x)+\cos(7x-5x)\right)=[/tex]

[tex]=\cos12x-\cos6x-\cos12x-\cos2x=-\cos6x-\cos2x=[/tex]

[tex]=-(\cos6x+\cos2x)=-2\cos\dfrac{6x+2x}{2} \cos\dfrac{6x-2x}{2} =-2\cos4x \cos2x=[/tex]

[tex]=-2(2\cos^22x-1) \cos2x=-2(2(2\cos^2x-1)^2-1) (2\cos^2x-1)[/tex]

Подставим известное значение косинуса [tex]\cos x=-\dfrac{\sqrt{3} }{3}[/tex]:

[tex]-2\cdot\left(2\cdot\left(2\cdot\left(-\dfrac{\sqrt{3} }{3}\right)^2 -1\right)^2-1\right)\cdot \left(2\cdot\left(-\dfrac{\sqrt{3} }{3}\right)^2-1\right)=[/tex]

[tex]=-2\cdot\left(2\cdot\left(2\cdot\dfrac{1}{3} -1\right)^2-1\right)\cdot \left(2\cdot\dfrac{1}{3}-1\right)=[/tex]

[tex]=-2\cdot\left(2\cdot\left(\dfrac{2}{3} -1\right)^2-1\right)\cdot \left(\dfrac{2}{3}-1\right)=-2\cdot\left(2\cdot\left(-\dfrac{1}{3} \right)^2-1\right)\cdot \left(-\dfrac{1}{3}\right)=[/tex]

[tex]=\dfrac{2}{3}\cdot\left(2\cdot\dfrac{1}{9} -1\right)=\dfrac{2}{3}\cdot\left(\dfrac{2}{9} -1\right)=\dfrac{2}{3}\cdot\left(-\dfrac{7}{9} \right)=-\dfrac{14}{27}[/tex]

Ответ: -14/27