The curves intersect when tan(5x) = 2sin(5x).
sin(5x)/cos(5x) = 2sin(5x)
sin(5x) = 2sin(5x)cos(5x)
sin(5x) - 2sin(5x)cos(5x) = 0
sin(5x)[1 - 2cos(5x)] = 0
sin(5x) = 0 or cos(5x) = 1/2
5x = kπ, 5x = π/3+2kπ, or 5x = 5π/3+2kπ
x = kπ/5, π/15+2kπ/5, or π/3+2kπ/5
If k=0, we have x = 0, π/15, π/3 (too large)
If k=-1. we have x = -π/5 (too small), π/3 (too large), -π/15
For all other values of k, the solutions lie outside of the interval
So, for the interval [-π/15, π/15], the only points of intersection occur when x = -π/15, 0, and π/15.
The area of the region between x=-π/15 and x=0 is the same as the area of the region between x=0 and x=π/15.
Between x=0 and x=π/15, 2sin(5x) ≥ tan(5x).
Area = 2∫(0 to π/15)[2sin(5x) - tan(5x)]dx
= 2[(-2/5)cos(5x)+(1/5)ln lcos(5x)l](from 0 to π/15)
= [(-4/5)cos(π/3) + (2/5)ln(cos(π/3)) + (4/5)cos0 - (2/5)ln(cos0)]
= -2/5 + (2/5)ln(1/2) + 4/5 - 0
= 2/5 - (2/5)ln2