Answer
$L\left( x \right) = - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6} + \frac{1}{2}$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \cos 2x,{\text{ }}a = \frac{\pi }{6} \cr
& {\text{Evaluate }}f\left( x \right){\text{ at }}a = \frac{\pi }{6} \cr
& f\left( {\frac{\pi }{6}} \right) = \cos 2\left( {\frac{\pi }{6}} \right) \cr
& f\left( {\frac{\pi }{6}} \right) = \cos \left( {\frac{\pi }{3}} \right) \cr
& f\left( {\frac{\pi }{6}} \right) = \frac{1}{2} \cr
& {\text{Differentiating the given function}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\cos 2x} \right] \cr
& f'\left( x \right) = - 2\sin 2x \cr
& {\text{Evaluate }}f'\left( x \right){\text{ at }}a = \frac{\pi }{6} \cr
& f'\left( {\frac{\pi }{6}} \right) = - 2\sin 2\left( {\frac{\pi }{6}} \right) \cr
& f'\left( {\frac{\pi }{6}} \right) = - 2\sin \left( {\frac{\pi }{3}} \right) \cr
& f'\left( {\frac{\pi }{6}} \right) = - \sqrt 3 \cr
& {\text{The linear approximation at }}f\left( a \right){\text{ is given by: }} \cr
& L\left( x \right) = f\left( a \right) + f'\left( a \right)\left( {x - a} \right){\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Substituting }}f\left( {\frac{\pi }{6}} \right) = \frac{1}{2},{\text{ }}f'\left( {\frac{\pi }{6}} \right) = - \sqrt 3 {\text{ and }}a = \frac{\pi }{6}{\text{ into }}\left( {\bf{1}} \right) \cr
& L\left( x \right) = \frac{1}{2} - \sqrt 3 \left( {x - \frac{\pi }{6}} \right) \cr
& {\text{Simplifying}} \cr
& L\left( x \right) = \frac{1}{2} - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6} \cr
& L\left( x \right) = - \sqrt 3 x + \frac{{\sqrt 3 \pi }}{6} + \frac{1}{2} \cr} $$