Question

помогите решить (тригонометрия)

Answer 1

1
4cos²x+2=5sin²x+3sinxcosx
4cos²x+2cos²x+2sin²x-5sin²x-3sinxcosx=0/cos²x
3tg²x+3tgx-6=0
tgx=a
3a²+3a-6=0
a²+a-2=0
a1=a2=-1 U a1*a2=-2
a1=-2⇒tgx=-2⇒x=-arctg2+πn,n∈z
a2=1⇒tgx=1⇒x=π/4+πn,n∈z
2
sinx+cosx=√2
√2(1/√2*sinx+1/√2*cosx)=√2
sin(x+π/4)=1
x+π/4=π/2+2πn
x=π/4+2πn,n∈z
3
5sin²x+3sinxcosx=4
5sin²x+3sinxcosx-4sin²x-4cos²x=0/cos²x
tg²x+3tgx-4=0
tgx=a
a²+3a-4=0
a1+a2=-3 U a1*a2=-4
a1=-4⇒tgx=-4⇒x=-arctg4+πn,n∈z
a2=1⇒tgx=1⇒x=π/4+πn,n∈z
4
sinxcos2x+cosxsin2x=√3/2
sin(x+2x)=√3/2
sin3x=√3/2
3x=(-1)^n*π/3+πn
x=(-1)^n*π/9+πn/3,n∈z
5
tgx+ctgx=2
tgx+1/tgx=2
tg²x-2tgx+1=0,tgx≠0
(tgx-1)²=0
tgx-1=0
tgx=1
x=π/4+πn,n∈z
6
cos5x+cos3x=0
2cos4x*cosx=0
cos4x=0⇒4x=π/2+πn⇒x=π/8+πn/4,n∈z
cosx=0⇒x=π/2+πn,n∈z
7
sin²x+2sinx+sinπ/2=0
sin²x+2sinx+1=0
(sinx+1)²=0
sinx+1=0
sinx=-1
x=-π/2+2πn,n∈z