1) Решить уравнение
cos4x = -2cos²x
2) Найти корни, принадлежащие заданному промежутку
[0;180⁰]
1) cos4x = -2cos²x
2cos²x - 1 = - 1 - cos2x
2cos²x - 1 + 1 + cos2x = 0
2cos²x + cos2x = 0
2cos²x +1 -2sin²x=0
2cos²x -2sin²x= - 1
2(cos²x - sin²x) = -1
cos²x - sin²x = -1/2
cos2x = -1/2
2x = ±2π/ 3 +2 πn, n ∈ Z
x = ±π/ 3 + πn, n ∈ Z
2) [0;180⁰]
x = -π/ 3 + πn, n ∈ Z
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Verified answer
1) cos4x = -2cos²x
2cos²x - 1 = - 1 - cos2x
2cos²x - 1 + 1 + cos2x = 0
2cos²x + cos2x = 0
2cos²x +1 -2sin²x=0
2cos²x -2sin²x= - 1
2(cos²x - sin²x) = -1
cos²x - sin²x = -1/2
cos2x = -1/2
2x = ±2π/ 3 +2 πn, n ∈ Z
x = ±π/ 3 + πn, n ∈ Z
2) [0;180⁰]
x = -π/ 3 + πn, n ∈ Z