cos2x + √2cos (п/2 + х) + 1= 0
cos2x - √2sinx + 1 = 0
1 – 2sin^2 x - √2sinx + 1 = 0
2sin^2 x + √2sinx–2 = 0
[sinx=(-√2 - 3√2) / 4; sinx = (-√2 +3√2) / 4
[sinx = -√2; sinx = √2/2
sinx=√2/2
[x = п/4 +2пn, n∈Z; x = 3п/4 + 2пn, n∈Z
А теперь корни, ∈ промежутку [2п; 7п/2]
2п + п/4 = 9п/4
3п – п/4 = 11п/4
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Answers & Comments
cos2x + √2cos (п/2 + х) + 1= 0
cos2x - √2sinx + 1 = 0
1 – 2sin^2 x - √2sinx + 1 = 0
2sin^2 x + √2sinx–2 = 0
[sinx=(-√2 - 3√2) / 4; sinx = (-√2 +3√2) / 4
[sinx = -√2; sinx = √2/2
sinx=√2/2
[x = п/4 +2пn, n∈Z; x = 3п/4 + 2пn, n∈Z
А теперь корни, ∈ промежутку [2п; 7п/2]
2п + п/4 = 9п/4
3п – п/4 = 11п/4