Определить количество корней уравнения 2 соs²(x/4+п/4)+6cos²(x/8+п/8)=2 на отрезке [0;12п]

Там есть решение

2 соs²(x/4+п/4)+6cos²(x/8+п/8)=2⇒2 соs²(x/4+п/4)+6cos²[(x/4+п/4)/2]=2⇒2 соs²(x/4+п/4)+3соs(x/4+п/4)+3=2⇒2 соs²(x/4+п/4)+3соs(x/4+п/4)+1=0соs(x/4+п/4)=t2t²+3t+1=0⇒t₁,₂=[-3+-√(9-8)]/4=(-3+-1)/4⇒t₁=-1  t₂=-1/21)cos(x/4+π/4)=-1⇒x/4+π/4=π+2πk⇒x/4=3π/4+2πk⇒x=3π+8πk  k∈Z2)cos(x/4+π/4)=-1/2⇒x/4+π/4=+-2π/3+2πl⇒x/4=+-2π/3-π/4+2πl⇒x=+-8π/3-π+8πl⇒x₁=-11π/3+8πl  x₂=5π/3+8πl  l∈Zx=3π;11π;7π/3;5π/3;29π/3n=5