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pestov2000
@pestov2000
July 2022
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помогите решить, срочно!
sin4x=3cos2x
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kratgv
Verified answer
sin4x=3cos2x
2·sin2x·cos2x -
3cos2x=0
cos2x·(2·sin2x - 3)=0
cos2x=0 или 2·sin2x - 3=0
2х=П/2+Пк, к∈Z 2·sin2x = 3
x=П/4+П/2·к, к∈Z sin2x =1,5
нет реш
Ответ:П/4+П/2·к, к∈Z
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Answers & Comments
Verified answer
sin4x=3cos2x2·sin2x·cos2x - 3cos2x=0
cos2x·(2·sin2x - 3)=0
cos2x=0 или 2·sin2x - 3=0
2х=П/2+Пк, к∈Z 2·sin2x = 3
x=П/4+П/2·к, к∈Z sin2x =1,5
нет реш
Ответ:П/4+П/2·к, к∈Z