2cos²x+5cosx+2=0 Пусть cosx=v ⇒ 2v²+5v+2=0 D=3 v₁=-2 v₂=-0,5
⇒ cosx=-2 x∉
cosx=-1/2 x=-π/3+2πn
Cos x - Sin x = Cos x - Cos(π/2 - x) =
= - 2Sinπ/4 * Sin(x-π/4) = -√2Sin(x-π/4)
14+6,3*(10^-3)=14(1+1,2+(10^-5t°)
1+1,2*(10^-5)t°=1,00045
1,2*(10^-5)t°=0,00045
t°=0,00045/1,2*100000
t°=37,5