Не получается sin²4x + cos²x = 2 sin 4x cos⁴x.

Уясни вопрос уравнения и я отвечу просто так не очень понятно

Объяснение:

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).