Решите уравнение

4sin^2x - 4cosx - 1 = 0

sin^2x - 0,5sin2x = 0

sin2x + sin6x = cos2x

1) 4sin²x-4cosx-1=0

4 (1-cos²x) - 4cosx-1=0

4-4cos²x-4cosx-1=0

4cos²x+4cosx-3=0

Пусть cosx=t, |t|≤1

4t²+4t-3=0

D=4²+4*4*3=64=8²

t₁ = (-4+8) / 8=1/2

t₂ = (-4-8) / 8=-1.5 <-1 не подходит по замене

cosx=1/2

x=+-π/6+2πn, n∈Z

2) sin²x-0.5*sin2x=0

sin²x-0.5*2sinx*cosx=0

sin²x-sinx*cosx=0

sinx (sinx-cosx) = 0

sinx=0

x=πn, n∈Z

sinx-cosx=0 |:cosx

tgx-1=0

tgx=1

x=π/4+πn, n∈Z

3) sin2x+sin6x=cos2x

2sin ((2x+6x) / 2) * cos ((6x-2x) / 2) = cos2x

2sin4x*cos2x=cos2x

2sin4x*cos2x-cos2x=0

2cos2x (sin4x-0.5) = 0

cos2x=0

2x=π/2+πn, n∈Z

x=π/4+π*n/2, n∈Z

sin4x=0.5

4x = (-1) ⁿ*π/6+πn, n∈Z

x = (-1) ⁿ*π/24+πn/4, n∈Z