Решите уравнение

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

 

sin^2x - 0,5sin2x= 0

 

sin2x + sin6x = cos2x

 

1) 4sin²x-4cosx-1=04(1-cos²x)-4cosx-1=04-4cos²x-4cosx-1=04cos²x+4cosx-3=0Пусть cosx=t, |t|≤14t²+4t-3=0D=4²+4*4*3=64=8²t₁=(-4+8)/8=1/2t₂=(-4-8)/8=-1.5 <-1 не подходит по заменеcosx=1/2x=+-π/6+2πn, n∈Z2)sin²x-0.5*sin2x=0sin²x-0.5*2sinx*cosx=0sin²x-sinx*cosx=0sinx(sinx-cosx)=0sinx=0x=πn, n∈Zsinx-cosx=0  |:cosxtgx-1=0tgx=1x=π/4+πn, n∈Z3) sin2x+sin6x=cos2x2sin((2x+6x)/2)*cos((6x-2x)/2)=cos2x2sin4x*cos2x=cos2x2sin4x*cos2x-cos2x=02cos2x(sin4x-0.5)=0cos2x=02x=π/2+πn, n∈Zx=π/4+π*n/2, n∈Zsin4x=0.54x=(-1)ⁿ*π/6+πn, n∈Zx=(-1)ⁿ*π/24+πn/4, n∈Z