Ответ:
a)π/2+2πn, n∈Z; –π/2+2πm, m∈Z; π/6+2πk; 5π/6+2πl, l∈Z.б)(5π)/2; (3π)/2; (13π)/6
Объяснение:
) 2·sin3x–2·sinx+cos2x=0
2·sin3x–2·sinx+1–sin2x=0
2·sinx(sin2x–1)–(–1+sin2x)=0
(sin2x–1)(2·sinx–1)=0
sin2x–1=0 или 2·sinx–1=0
sin2x=1
1)sinx=1
х=π/2+2πn, n∈Z
2)sinx=–1
х=–π/2+2πm, m∈Z
3)2·sinx–1=0
2·sinx=1
sinx=1/2
x=π/6+2πk, k∈Z или х=5π/6+2πl, l∈Z.
б)1)π⩽π/2+2πn⩽(5π)/2
1/2⩽2n⩽2
1/4⩽n⩽1
n=1
x=π/2+2π=(5π)/2
2)π⩽ –π/2+2πm ⩽(5π)/2
3/2⩽2m⩽3
3/4⩽m⩽3/2
m=1
x= –π/2+2π=(3π)/2
3)π⩽ π/6+2πk ⩽(5π)/2
5/6⩽ 2k ⩽14/6
5/12⩽ k ⩽14/12
k=1
x=π/6+2π=(13π)/6
4)π⩽ 5π/6+2πl ⩽(5π)/2
1/6⩽ 2l ⩽10/6
1/12⩽ l ⩽10/12
l=ø