Answer
$ \approx 0.614$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 2\sin x,{\text{ }}g\left( x \right) = \tan x \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval 0}} \leqslant x \leqslant \frac{\pi }{3} \cr
& g\left( x \right) \geqslant f\left( x \right){\text{ on the intervals }} - \frac{\pi }{3} \leqslant x \leqslant 0{\text{ }} \cr
& {\text{Using the symmetry we can define the area enclosed as:}} \cr
& A = \,2\int_0^{\pi /3} {\left( {2\sin x - \tan x} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = \,2\left[ { - 2\cos x + \ln \left| {\cos x} \right|} \right]_0^{\pi /3} \cr
& A = \,2\left[ { - 2\cos \left( {\frac{\pi }{3}} \right) + \ln \left| {\cos \left( {\frac{\pi }{3}} \right)} \right|} \right] - 2\left[ { - 2\cos \left( 0 \right) + \ln \left| {\cos \left( 0 \right)} \right|} \right] \cr
& A = \,2\left[ { - 1 + \ln \left( {\frac{1}{2}} \right)} \right] - 2\left[ { - 2 + 0} \right] \cr
& A = - 2 + 2\ln \left( {\frac{1}{2}} \right)\, + 4 \cr
& A = 2 + 2\ln \left( {\frac{1}{2}} \right) \cr
& A = 2 - 2\ln \left( 2 \right) \cr
& A \approx 0.614 \cr} $$
