Решить однородные уравнения: 1)5cos2x-3sin2x=0.2)cos^2x-3sin^2x=0.3)3sin^2x-4sinx*cosx+5cos^2=2.4)sin^4x-cos^4x=sin2x.5)2sin^2-cos(pi/2+x)sin(3/2pi+x)-sin^2(3pi/2+x)=4arccos1. решить нужно

1)5cos2x-3sin2x=0
5сos²x-5sin²x-6sinxcosx=0/-cos²x≠0
5tg²x+6tgx-5=0
tgx=a
5a²+6a-5=0
D=36+100=136
a1=(-6-2√34)/10=-0,6-0,2√34⇒tgx=-0,6-0,2√34⇒x=-arctg(0,6+0,2√34)+πn
a2=-0,6+0,2√34⇒tgx=0,2√34-0,6⇒x=arctg(0,2√34-0,6)+πn
2)cos^2x-3sin^2x=0
(1+cos4x)/2-3(1-cos4x)/2=0
1+cos4x-3+3cos4x=0
4cos4x=2⇒cos4x=1/2⇒4x=+-π/3+2πn⇒x=+-π/12+πn/2
3)3sin^2x-4sinx*cosx+5cos^2x=2
3sin^2x-4sinx*cosx+5cos^2x-2sin²x-2cos²x=0
sin²x-4sinxcosx+3cos²x=0 /cos²x≠0
tg²x-4tgx+3=0
tgx=a
a²-4a+3=0⇒a1+a2=4 U a1*a2=3
a1=1⇒tgx=1⇒x=π/4+πn
a2=3⇒tgx=3⇒x=arctg3+πn
4)sin^4x-cos^4x=sin2x
(sin²x-cos²x)(sin²x+cos²x)=sin2x
sin²x-cos²x-2sinxcosx=0 /cos²x≠0
tg²x-2tgx-1=0
tgx=a
a²-2a-1=0
D=4+4=8
a1=(2-2√2)/2=1-√2⇒tgx=1-√2⇒x=arctg(1-√2)+πn
a2=1+√2⇒tgx=1+√2⇒x=arctg(1+√2)+πn
5)2sin^2x-cos(pi/2+x)sin(3/2pi+x)-sin^2(3pi/2+x)=4arccos1
 2sin²x-(-sinx)*(-cosx)-cos²x=4*0
2sin²x-sinxcosx-cos²x=0/cos²x≠0
2tg²x-tgx-1=0
tgx=a
2a²-a-1=0
D=1+8=9
a1=(1-3)/4=-1/2⇒tgx=-1/2⇒x=-arctg1/2+πn
a2=(1+3)/4=1⇒tgx=1⇒x=π/4+πn