Question

Решите уравнение4sin^2x - 4cosx - 1= 0sin^2x - 0,5sin2x= 0sin2x + sin6x = cos2x

Answer 1

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