PhysMaths
1) sin x = cos x равны только в точках π/4 и 5π/4 x = π/4 + 2πk, k ∈ Z x = 5π/4 + 2πn, n ∈ Z 2)1 - cos2x = sinx 1 - (1-2sin^2 x) = sin x 1 - 1 + 2sin^2 x = sinx 2sin^2x - sinx = 0 sinx*(2sinx-1)=0 1. sin x = 0 x = πk, k ∈ Z 2. 2sin x - 1 = 0 2sin x = 1 sinx = 1/2 Синус равен 1/2 в точках π/6 и 5π/6 x1 = π/6 + 2πn, n ∈ Z x2 = 5π/6 + 2πm, m ∈ Z
Answers & Comments
x = π/4 + 2πk, k ∈ Z
x = 5π/4 + 2πn, n ∈ Z
2)1 - cos2x = sinx
1 - (1-2sin^2 x) = sin x
1 - 1 + 2sin^2 x = sinx
2sin^2x - sinx = 0
sinx*(2sinx-1)=0
1. sin x = 0
x = πk, k ∈ Z
2. 2sin x - 1 = 0
2sin x = 1
sinx = 1/2
Синус равен 1/2 в точках π/6 и 5π/6
x1 = π/6 + 2πn, n ∈ Z
x2 = 5π/6 + 2πm, m ∈ Z