Question

2cosx-sinx-1=0 срочно!!!!

Answer 1

2cosx - sinx - 1 = 0
2cos³(x/2) - 2sin²9x/) - 2sin(x/2)cos(x/2) - sin³(x/) - cos²(x/2) = 0
cos²(x/) - 2sin(x/2)cos(x/2) - 3sin²(x/2) = 0 делим на ( cos³x/2 ≠ 0)
3tg³(x/2) + 2tg(x/2) - 1 = 0
D = 4 + 4*3*1 = 16
1)  tg(x/2) = (- 2 - 4)/6
tg(x/2) = - 1
x/2 = - π/4 + πk, k∈Z
x₁ = - π/2 + 2πk, k∈Z
2)  tg(x/2) = ( - 2 + 4)/6
tg(x/2) = 1/3
x/2 = arctg(1/3) + πn, n∈Z
x₂ = 2*arctg(1/3) + 2πn, n∈Z