Sin^2x/cos^2x-3/cosx+3=0 (cosx≠0, x≠pi/2+2pi*k, k∈Z)
cosx=t
(1-t^2)/t^2-3/t+3=0
(1-t^2-3t+3t^2)/t^2=0
(2t^2-3t+1)/t^2=0
2t^2-3t+1=0
D=9-8=1
t1=(3+1)/4=1
t2=(3-1)/4=1/2
cosx=1 cosx=1/2
x=2pn, n∈Z x=pi/3+2pi*n, n∈Z x= -pi/3+2pi*n, n∈Z
2) корни из отрезка [-3п;-3п/2]
x1= -2p
x2= -7pi/3
x3= -5pi/3
x4= -8pi/3