5sin^2(x)+8cos(x)=8
8-5sin^2(x)-8cos(x)=0
2.5-5sin^2(x)+5.5-8cos(x)=0
5cos^2(x)+3-8cos(x)=0
cos(x)=(8+-sqrt(64-60))/10=(8+-2)/10= 1 или 0,6
Значит:
x = 2 π n, n ∈ Z
x = 2 π k - arccos(3/5), k ∈ Z
x = 2 π k + arccos(3/5), k ∈ Z
Но sin(x)>0
Тогда:
x = 2 π k + arccos(3/5), k ∈ Z