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
