Given:
#-2/5*cos(4pi*x)=cos(10pi*x)#
#-2cos(4pi*x)=5cos(10pi*x)#
Let
#2pix=t#
#-2cos2t=5cos5t#
#cos2t=cos^2t-sin^2t#
wkt
#(cos5t+isin5t)=(cost+isint)^5#
#=cos^5t+5cos^4tisint+10cos^3ti^2sin^2t+10cos^2ti^3sin^3t+5costi^4sin^4t+i^5sin^5t#
#=cos^5t+5icos^4tsint-10cos^3tsin^2t-10icos^2sin^3t+5costsin^4t+isin^5t#
Rearranging into real and imaginary parts
#cos5t+isin5t=(cos^5t-10cos^3tsin^2t+5costsin^4t)#
#+i(5cos^4tsint-10cos^2sin^3t+sin^5t#
For cos5t, equate real parts
#cos5t=cos^5t-10cos^3tsin^2t+5costsin^4t#
We have
#-2cos2t=5cos5t#
Thus,
#-2(cos^2t-sin^2t)=5(cos^5t-10cos^3tsin^2t+5costsin^4t)#
#2(cos^2t-sin^2t)+5(cos^5t-10cos^3tsin^2t+5costsin^4t)=0#
#2cos^2t-2sin^2t+5cos^5t-50cos^3tsin^2t+25costsin^4t=0#
#sin^2t=1-cos^2t#
#2cos^2t-2(1-cos^2t)+5cos^5t-50cos^3t(1-cos^2t)+25cost(1-cos^2t)^2=0#
#2cos^2t-2+2cos^2t+5cos^5t-50cos^3t+50cos^5t+25cost(1-2cos^2t+cos^4t)=0#
#-2+4cos^2t-50cos^3t+55cos^5t+25cost-50cos^3t+25cos^5t=0#
u=cost
#-2+4u^2-50u^3+55u^5+25u-50u^3+25u^5=0#
#(55+25)u^5-(50+50)u^3+4u^2+25u-2=0#
#80u^5-100u^3+4u^2+25u-2=0#
which is a 5th degree polynomial in u having five maximum roots which can be found by trial and error
