(2sin2x-sin4x)/(sin4x+2sin2x)=(2sin2x-2sin2x·cos2x)/(2sin2x·cos2x+2sin2x)=
=[2sin2x(1-cos2x)]/[2sin2x·(cos2x+1)]=(1-cos2x)/(1+cos2x)=
=(cos²x+sin²x-cos²x+sin²x)/(cos²x+sin²x+cos²x-sin²x)=
=2sin²x/2cos²x=2tg²x
Sin Фи = -1
Фи = -π/2 +2πn, n∈Z
Фи∈[0; 4π]
0≤ -π/2+2πn ≤ 4π
π/2 ≤ 2πn ≤ 4π +π/2
π/2 ≤ 2πn ≤ 9π/2
<u> π </u> ≤ n ≤ <u> 9π </u>
2*2π 2*2π
1/4 ≤ n ≤ 9/4
0.25 ≤ n ≤ 2.25
n=1 Фи= -π/2 +2π = 3π/2
n=2 Фи= -π/2 +2π*2 = -π/2 +4π = 7π/2
Ответ: 3π/2; 7π/2.
F(x)=3x+tgx
F(π/6)=3π/6+tgπ/6=π/2+(√3)/3
5ax^2-10ax-yx+2y-x+2=(5ax^2-10ax)-(yx-2y)-(x-2)=5ax(x-2)-y(x-2)-(x-2)=(x-2)(5ax-y-1). Готово)