Помогите решить. √2cosx-1=0
По дате
По Рейтингу
√2·cos x - 1 = 0;
√2·cos x = 1;
cos x = 1/√2;
x = ±arccos(1/√2) + 2πk = ±π/4 + 2πk, k∈ℤ.
√2cosx-1=0
√2cosx=1
cosx=1/√2
x=arccos(1/√2)
√2·cos x - 1 = 0;
√2·cos x = 1;
cos x = 1/√2;
x = ±arccos(1/√2) + 2πk = ±π/4 + 2πk, k∈ℤ.
√2cosx-1=0
√2cosx=1
cosx=1/√2
x=arccos(1/√2)