Решите уравнение
1) sin 2x=корень из 3 cos x
2) sin 2x=корень из 2 cos x
1) 2 sin x cos x - √3 cos x = 0
cos x (2 sin x - √3) = 0
cos x = 0 2 sin x - √3 = 0
x₁ = π/2 + πn, n∈Z sin x = √3/2
x₂ = (-1)^n · π/3 + πn, n∈Z
2) 2 sin x cos x - √2 cos x = 0
cos x (2sin x - √2) = 0
cos x = 0 2 sin x - √2 = 0
x₁ = π/2 + πn, n∈Z sin x = √2/2
x₂ = (-1)^n · π/4 + πn, n∈Z
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1) 2 sin x cos x - √3 cos x = 0
cos x (2 sin x - √3) = 0
cos x = 0 2 sin x - √3 = 0
x₁ = π/2 + πn, n∈Z sin x = √3/2
x₂ = (-1)^n · π/3 + πn, n∈Z
2) 2 sin x cos x - √2 cos x = 0
cos x (2sin x - √2) = 0
cos x = 0 2 sin x - √2 = 0
x₁ = π/2 + πn, n∈Z sin x = √2/2
x₂ = (-1)^n · π/4 + πn, n∈Z