Question

cos^2x+4sin^2x=2sin2x

Answer 1

cos²x+4sin²x=2sin2x
cos²x+4sin²x-2sin2x=0
cos²x+4sin²x-2*2sinx*cosx=0 |:cos²x≠0
1+4tg²x-4tgx=0
a=tgx
4a²-4a+1=0
D=(-4)²-4*4*1=16-16=0
a(₁,₂)= 4/2*4=0,5
tgx=0,5
x=arctg0,5 +πn, n∈Z