(36^sinx)^cosx = 6^V2sinx
(6^2sinxcosx)=6^V2sinx
2sinxcosx=V2sinx
2cosx=V2
cosx=V2/2
x=+- p/4 + 2pk
а)
или
б) На [2П; 7П/2]:
1)
2)
ответ: 2П; 3П; 9П/4.
(36^sinx)^cosx = 6^V2sinx
(6^2sinxcosx)=6^V2sinx
2sinxcosx=V2sinx
2cosx=V2
cosx=V2/2
x=+- p/4 + 2pk
а)
б) На [2П; 7П/2]:
1)
2)
ответ: 2П; 3П; 9П/4.