4*(1+cos2x)²/4 -1 =cos2x
(1+cos2x)² -(1+cos2x)=0
(1+cos2x)(1+cos2x-1)=0
(1+cos2x)=0⇒cos2x=-1⇒2x=π+2πn⇒x=π/2+πn⇒x=-π/2∈[-5π/6;π/6]
(1+cos2x-1)=0⇒cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2⇒x=-3π/4;-π/4∈[-5π/6;π/6]
1. y=log₂(2x+3)
2.y=1/3cos(3x-π/2)-π³-e², x₀=π/3
y'=1/3 (-sin(3x-π/2))*3=-sin(3x-π/2)
y'(x₀)=-sin(3*π/3-π/2)=-sin(π-π/2)=-sin(π/2)=-1
3. y=x², x₀=0,25
y'=2x
k=y'(x₀)=2*0,25=0,5
4. sin(π/6-3x)-1/2=0
sin(π/6-3x)=1/2
π/6-3x=π/2+2πn
3x=π/6-π/2+2πn=π/6-3π/6+2πn=-2π/6+2πn=-π/3+2πn
x=-π/9+2πn/3 , n∈Z
5в2с и -5в2с сокращаются. -4с*3в-4с*(-2) Вынесем -4с. И получим -4с(3в-2)
Итак, первое число равен 2, тогда второе - (2+2=4), а третье - (2+2*2)=6
-5×-7=0 -8×+9=-7 ×+12=3×4/
-5×=7 -8×=-7-9 ×+3×=12
×=7÷5 -8×=-16 4×=12
×=1.4 ×=2 ×=3
г) -4 (×+2)+3 (×-1)-2=4 (×-2)+9
-4×+(-6)+3×-3-2=4×-8+9
-6+3×-5=1
3×-5-6=1
3×-11=1
3×=1+11
3×=12
×=4