3sin²x-cosx+1=0
3(1-cos²x)-cosx+1=0
3cos²x+cosx-2=0
cosx=t,, t∈[-1;1]
3t²+t-2=0
t₁=-1, t₂=2/3
1. cosx=-1, <u>x₁=π+πn, n∈Z</u>
2. cosx=2/3 <u>x₂=+-arccos(2/3)+2πn, n∈Z </u>
Y=f(x0)+f'(x0)(x-x0)
f(x0)=2
f'(x)=-4/x^2
f'(x0)=-1
y=2-1*(x-2)
y=2-x+2
y=-x
(10/5)а-6/5-5,2+а=2а+а-5,2-1,2=3а-4=3*(2/3)-4=2-4=-2
1) 1.5х в квадрате-х-2=1
1.5х-2=1
1.5х=1+2
1.5=3
х=3:1.5
х=2
1 2 9+3 1 3=1 2 9+ 3 3 9=4 5 9
5 3 4-2=3 3 4
4-1 7 9=3 7 9
4 1 3-1 1 2=4 2 6-1 3 6=3 8 6- 1 3 6=2 5 6