1)кос2х+синх=0 2)син2х-косх=2синх-1

Question

1)кос2х+синх=0
2)син2х-косх=2синх-1

Answer 1

1)cos2x-sinx=0
cos²x-sin²x-sinx=0
1-2sin²x-sinx=0
2sin²x+sinx-1=0
D=1+8=9
sinx1=(-1+3)/4 ;sinx2=(-1-3)/4
sinx1=1/2; sinx2=-1
x1=(-1)^(n+1)π/6+πn, n∈Z
x2=-π/2+2πn, n∈Z
2)sin2x-cosx