Sin(2x)+cos^2(x)=0 3cos^2(x)+8sinx*cosx-4=0 решить !

2sinxcosx+cos^2x=0
cosx(2sinx+cosx)=0
cosx=0, x=pi/2+pi*n
2sinx=-cosx|:cosx≠0
2tgx=-1
tgx=-1/2
x=arctg(-1/2)+pi*k

3cos^2x+8sinxcosx-4=0
3cos^2x+8sinxcosx-4(sin^2x+cos^2x)=0
8sinxcosx-4sin^2x-cos^2x=0|:cos^2x≠0
8tgx-4tg^2x-1=0
4tg^2x-8tgx+1=0
D=8^2-4*4=20
√D=2√5
tgx=(8+2√5)/8=1+√5/4
tgx=1-√5/4
x=arctg(1+√5/4)+pi*k
x=arctg(√5/4-1)+pi*n