Помогите! 4^cosx+4^-cosx=5\2 [-3П ; -3П\2]

$4^{cosx}+4^{-cosx}=5/2\\\\4^{cosx}+ \frac{1}{4^{cosx}}=5/2\\\\t=4^{cosx}\\\\t+ \frac{1}{t}= \frac{5}{2}|*2t\\\\2t^2-5t+2=0\\D=(-5)^2-4*2*2=25-16=9=3^2\\t_1=(5+3)/4=2\\t_2=(5-3)/4=1/2$

$4^{cosx}=2\\2^{2cosx}=2^1\\2cosx=1\\cosx=1/2\\x_1=б \pi /3+2 \pi n, n\in Z\\\\4^{cosx}=1/2\\2^{2cosx}=2^{-1}\\2cosx=-1\\cosx=-1/2\\x_2=б2 \pi /3+2 \pi n, n\in Z$

[-3π;-3π/2]
x₁=-3π+π/3=(-9π+π)/3=-8π/3
x₂=-2π-π/3=(-6π-π)/3=-7π/3
x₃=-2π+π/3=(-6π+π)/3=-5π/3