2cos^2(x/4) + 5sin(x/4) - 4 = 0
2(1-sin^2(x/4))+5sin(x/4) - 4 = 0
-2sin^2(x/4)+5sin(x/4) - 2 = 0 /:(-1)
2sin^2(x/4)-5sin(x/4) +2 = 0
sin(x/4)=t, t∈[-1; 1]
2t^2-5t+2=0
D=25-16=9>0
t=(5+3)/4=8/4=2⇒нет реш.
t=(5-3)/4=1/2
обратная подстановка
sin(x/4)=1/2
x/4=(-1)^k*pi/6+pin, n∈Z
x=(-1)^k*2pi/3+4pin, n∈Z