1) 2cos x = √3;

2) sin x = 1/2;

3) cos 2x = - 1

Question

Алгебра

Андорей

19 июля, 16:00

1) 2cos x = √3;

2) sin x = 1/2;

3) cos 2x = - 1

+1

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Оган · Answer 1

1) 2cox = √3

cosx = √3/2

x = + - arccos (√3/2) + 2πk, k ∈ Z

x = + - π/6 + 2πk

2) sin x = 1/2

x = (-1) ^k arcsin (1/2) + πk, k ∈ Z

x = (-1) ^k π/6 + πk

3) cos2x = - 1

2x = π + 2πk, k ∈ Z

x = π/2 + πk

1) 2cos x = √3;2) sin x = 1/2;3) cos 2x = - 1

1) 2cos x = √3;

2) sin x = 1/2;

3) cos 2x = - 1