First, let's convert #pi/3# to degrees. We do this with the conversion factor of #180/pi#

#pi/3 xx 180/pi = 60^@#

Now we have to calculate #cos60^@#. By the special triangle of sides #1, sqrt(3) and 2# we have that #cos60^@ = 1/2#

We can state the new expression as being #tan^-1(-2 xx 1/2) = tan^-1(-1)#

We must calculate the #tan^-1(-1)#. We can do this by first looking in which quadrants tan will be negative. Tan is negative in quadrants #II and IV#.

Furthermore, for the ratio to be #-1/1#, the #45^@, 45^@, 90^@; 1, 1, sqrt(2)# special triangle will be implicated.

By reference angles, we have that #tan^-1(-1) = 180 - 45^@ and 360 - 45^@#, so #tan^-1(-1) = 135^@ and 315^@#.

Depending on how your teacher wants your answer #tan^-1(-2cos(pi/3)) = 135^@ and 315^@ or tan^-1(-2cos(pi/3)) = (3pi)/4 and (7pi)/4#

Hopefully this helps!