Question

Помогите решить!
sin^2 2a=

Answer 1

Sin² 2α=1 - cos² 2α=1 - (1/4)²=5/16

Answer 2

Sin²2a=0
Sina=2sina*cosa
2sin²x*cos²x=0
sin²x=0
sin²a=1-cos2x/2=0
1-cos2x=0
-cos2x=-1
cos2x=1
cosx=a
x=±arccosa+2Пn, n$Z
cos2x=1
2x=±0+2Пn, n$Z
x=Пn, n$Z
Ответ :x=Пn, n$Z