Sinx=0,5√2=(√2)/2 ⇒ x=π/2+2πn, n∈Z
2sin²x=cosx+1
2(1-cos²x)-cosx-1=0
cosx=t
2t²-t-1=0
t=1 или t= -0,5
x=2πn или x=2π/3
sin²x-2sinxcosx-3cos²x=0
sin²x/cos²x-2sinxcosx/cos²x-3cos²x/cos²x=0
tg²x-2tgx-3=0
tgx= -1 или tgx=3
x= -π/4+2πn или x=arctg(3)+2πn
3sin2x+4cos2x=5
0,6sin2x+0,8cos2x=1
cosα=0,6 sinα=0,8
cosαsin2x+sinαcos2x=1
sin(2x+α)=1
2x+α=π/2
x=(π/2-α)/2+2πn
tgx≥-1
x≥-π/4
sinx<0,5<br>x∈(-7π/6;π/6)