Задача 55795 ...

5fbddb8164081d1dafcb1b94

25.11.2020 07:20:38

cos10x cos6x-cos²8x=0

5f3ea7e3faf909182968ddd9

25.11.2020 11:34:25

★

Формулы:
[r]cos α *cos β [/r]

[r]2cos^2 α =1+cos2 α [/r]

(cos16x+cos4x)/2- (1+cos16x)/2=0

cos4x-1=0

cos4x=1

4x=2πk, k ∈ Z

[b]x=(π/2)*k, k ∈ Z -[/b] о т в е т