См. вложение
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Решение
б) f(x)=2x+cos(4x-пи) = 2x - cos4x
f `(x) = 2 + 4sin4x
1) f `(x) = 0
<span>2 + 4sin4x = 0
</span>4sin4x = - 2
sin4x = - 1/2
4x = (-1)^(n) arcsin(-1/2) + πk, k ∈ Z
4x = (-1)^(n+1) arcsin(1/2) + πk, k ∈ Z
4x = (-1)^(n+1) (π/6)<span> + πk, k ∈ Z
</span>x = (-1)^(n+1) (π/24)<span> + πk/4, k ∈ Z
</span>2) f `(x) > 0
<span>2 + 4sin4x > 0
</span><span>sin4x > - 1/2
</span>arcsin(- 1/2) + 2πn < 4x < π - arcsin(-1/2) + 2πn, n ∈ Z
- π/6 + 2πn < 4x < π + π/6 <span>+ 2πn, n ∈ Z
</span>- π/24 + πn/2 < x < 7π/24 + πn/2, n ∈ Z
в) f(x) = cos2x
f `(x) = - 2sin2x
1) f `(x) = 0
<span>- 2sin2x = 0
</span>sin2x = 0
2x = πk, k ∈ Z
x = <span>πk/2, k ∈ Z
2) </span><span>- 2sin2x > 0
sin2x < 0
- </span>π - arcsin0 + 2πn < 2x < arcsin0 + 2πn, n ∈ Z
- <span>π + 2πn < 2x < 2πn, n ∈ Z
</span>- <span>π/2 + πn < x < πn, n ∈ Z</span>
1. 3x3-12x=0 2.16x2+4x=0
3x(x2-4)=0 4x(4x+1)=0
3x=0 или x2-4=0 4x=0 или 4x+1=0
x=0 или x2=4 x=0 или 4x=-1
x=2 x=-0,25
Ответ:x=0,x=2 Ответ:x=0,x=-0,25
3tg(x/3+pi/6)=3
tg(x/3+pi/6)=3/3
tg(x/3+pi/6)=1
x/3+pi/6=arctg 1 + pi n,n£Z
x/3=-pi/6 +pi/4+pi n,n£Z
x=-pi/3+pi/2+3pi n,n£Z
x= -2pi/6+3pi/6+3pi n ,n£Z
x=pi/6 + 3pi n ,n £Z
3-y+2(5-y)=4y-8
3-y+10-2y=4y-8
-7y=-21 (делим на минус 7)
y=3
