2) \(\dfrac{6\cos207^\circ}{\cos27^\circ}\)

Используем формулу \(\cos(180^\circ+x)=-\cos x\): \(\cos207^\circ=\cos(180^\circ+27^\circ)=-\cos27^\circ\).

Тогда \(\dfrac{6\cos207^\circ}{\cos27^\circ}=\dfrac{-6\cos27^\circ}{\cos27^\circ}=-6\).

Ответ: \(-6\).