функція у=f(x) парна.На проміжку (-∞;-3)вона спадає , а на проміжку (-2;0) зростає .Якою вона є на решті області визначення

The given equation can be re-written as sin

2

4x−2sin4xcos

4

x+cos

2

x=0

Add and subtract cos

8

x

∴(sin4x−cos

4

x)

2

+cos

2

x(1−cos

6

x)=0

Since both the terms are +ive (cos

6

x≤1), above is possible only when each term is zero for the same value of x.

sin4x−cos

4

x=0 .(1)

and cos

2

x(1−cos

6

x)=0 .(2)

From (2) cosx=0 or cos

2

x=1

∵z

3

=1⇒z=1 only

as other values will not be real.

Case I: If cosx=0 i.e., x=(n+

2

1

)π, then from (1)

sin4(n+

2

1

)π+0=0

or sin(4n+2)π=0 which is true.

∴x=(n+

2

1

)π (3)

Case II: When cos

2

x=1 i.e., sinx=0

∴x=rπ then from (1), sin4rπ−1=0 or −1=0 which is not true. Hence the only solution is given by (3).

1) 3х²-27=0

3х²=27

х²=9

х¹=3;х²=-3

2) 2х²+3х-5=0

а=2, b=3, c=-5

D=b²-4ac=3²-4×2×(-5)=9+40=49=9²

x½=-b±D/2a=-3±9/4=-3 и 1,25

3)(х+3)²-5х(х-2)=10(2х+1)

х²+6х+9-5х²+10=20х+10

х²+6х+9-5х²+10-20х-10=0

-4х²-14х+9=0/:-1

4х²+14х-9=0

а=4, b=14, c=-9

D=14²-4×4×(-9)=169+144=313=√313

x½=-14±√313/8

4) (x+2)²=43-6x

x²+4x+4-43+6x=0

x²+10x-39=0

a=1, b=10, c=-39

D=10²-4×1×(-39)=100+156=256=16²

x½=-10±16/2=3 и -13

5) (х-2)²+24=(2+3х)²

х²-4х+4+24=4+12х+9х²

х²-4х+4+24-4-12х-9х²=0

-8х²-16х+24=0/:-1

8х²+16х-24=0

a=8, b=16, c=-24

D=16²-4×8×(-24)=256+768=1024=32²

x½=-16±32/16=-3 и 1