Question

решите уравнение : √2(sinx+cosx)=4sinxcosx

Answer 1

Представим sinx+cosx, как сумму sinx+cosy
√2*(2*sin(П/4)*сos(x-П/4)=4*sinx*cosx
(√2*2*√2/2)*cos(x-П/4)=4*sinx*cosx
2*cos(x-П/4)=4*sinx*cosx
cos(x-П/4)=2*sinx*cosx
cos(x-П/4)=sin2*x
cos(x-П/4)=сos(П/2-2*x)
cos(П/2-2*x)-cos(x-П/4)=0
2*sin((П/2-2*x+x-П/4)/2)*sin(x-П/4-П/2+2*x)/2=0
sin((П/4-x)/2)*sin((3*x-3*П/4)/2)=0
sin(П/8-x/2)*sin(3*x/2-3*П/4)=0
sin(П/8-x/2)=0   П/8-x/2=П*n   x/2=П/8-П*n  x=П/4-2*Пn=П/4+2*П*n, nЄZ
sin(3*x/2-3*П/8)=0   3*x/2-3*П/8=П*k   3*x/2=3*П/8+П*k    3*x=3*П/4+2*П*k
x=П/4+2*П*k/3, kЄZ

Answer 2

cosα =cosβ не означает α =β . например ; cos7π/3 =cos17π/3

Answer 3

x=π/4-2*πn , n ∈ Z OK ! || x=π/4-2*πn =π/4+2*πk, k ∈ Z