1
а)√3tgx-1=0
tgx=1/√3
x=π/6+πn,n∈Z
б) 2sin(-x/2) = 1
sin(x/2)=-1/2
x/2=(-1)^(n+1)*π/6+πn,n∈Z
x=(-1)^(n+1)*π/3+2πn,n∈Z
в) 2cos(2x+π/4) = -√2
cos(2x+π/4)=-√2/2
2x+π/4=+-3π/4+2πn,n∈Z
2x=-π+2πn,n∈Z U 2x=π/2+2πn,n∈Z
x=-π/2+πn,n∈Z U x=π/4+πn,n∈Z
г) sin (x-π/6) = -√3/2
x-π/6=(-1)^(n+1)*π/3+πn,n∈Z
x=π/6+(-1)^(n+1)*π/3+πn,n∈Z
2
д) 6cos²x + 7cosx - 8 = 0
cosx=a
6a²+7a-8=0
D=49+192=241
a1=(-7-√241)/12⇒cosx=(-7-√241)/2<-1 нет решения<br>a2=(-7+√241)/12⇒cosx=(√241-7)/12⇒x=+-arccos(√241-7)/12+2πn,n∈Z
x=-arccos(√241-7)/12∈(-π;π/2)
x=arccos(√241-7)/12∈(-π;π/2)
e) sinxcosx - cos²x=0/cos²x≠0
tgx-1=0
tgx=1
x=π/4+πn,n∈Z
x=π/4(-π;π/2)
ж) 3tg²2x - 2ctg (π/2 + 2x) -1 = 0
3tg²2x+2tg2x-1=0
tg2x=a
3a²+2a-1=0
D=4+12=16
a1=(-2-4)/6=-1⇒tg2x=-1⇒2x=-π/4+πn,n∈Z⇒x=-π/8+πn/2,n∈Z
x=-π/8(-π;π/2)
a2=(-2+4)/6=1/3⇒tg2x=1/3⇒x=1/2arctg1/3+πn,n∈Z
x=1/2arctg1/3(-π;π/2)
з) 5cos²α - sinxcosx = 2
5cos²x-sinxcosx-2sin²x-2cos²x=0
2sin²x+sinxcosx-3cos²x=0/cos²x≠0
2tg²x+tgx-3=0
tgx=a
2a²+a-3=0
D=1+24=25
a1=(-1-5)/4=-1,5⇒tgx=-1,5⇒x=-arctg1,5+πnx=-arctg1,5∈(-π;π/2)
a2=(-1+5)/4=1⇒tgx=1⇒x=π/4+πn,n∈Z
x=π/4∈(-π;π/2)
x=-3π/4∈(-π;π/2)