2 cos^3 x/5 + sin^2 x/5=1
так как
sin²(x/5)=1-cos²(x/5)
получаем кубическое уравнение
2 cos³(x/5)+1-cos²(x/5)=1
2cos³(x/5)-cos²(x/5)=0
cos²(x/5)(2cos(x/5)-1)=0
cos(x/5)=0 или 2 cos(x/5)-1=0
x/5 = (π/2) + πn,n∈Z cos(x/5)=1/2
x = (5π/2) + 5πn,n∈Z x/5= ± arccos(1/2) + 2πk,k∈Z
x/5= ± (π/3) + 2πk,k∈Z
x= ± (5π/3) + 10πk,k∈Z
Ответ. x = (5π/2) + 5πn,n∈Z ; x= ± (5π/3) + 10πk,k∈Z