Alternate Method
I think the question is using known trigonometric identities.
As #pi/3*1/3=pi/9# we can compute #sec (pi/9)# using #cos (pi/3)#.
The method is as given below
# cos (pi/3) = 1/2#
# cos(3 pi/9) = 1/2#
# cos 3theta = cos(2theta +theta)#
# = cos2thetacostheta-sin2thetasintheta#
#=2cos^2thetacostheta-costheta-2sin^2thetacostheta#
#=4cos^3theta-3costheta#
If #theta = pi/9# and we assume #x=costheta# then we have
#4x^3-3x-1/2 = 0#
#8x^3-6x-1 = 0#
Solving the equation results in 3 roots.
#x=costheta=( -0.76604, -0.17365, 0.93969)# as the angle is less than #pi/2# cosine is positive. Hence choose the positive root.
#cos(pi/9) = 0.93969#
#sec (pi/9) = 1/cos(pi/9) = 1/0.93969#
#sec(pi/9) = 1.0642#
