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furman1995
@furman1995
June 2022
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2cos²x=√3sin(3П/2+x)
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2cos²x=√3sin(3π/2+x)
2cos²x= -√3cosx
2cos²x+√3cosx=0
cosx(2cosx+√3)=0
cosx=0 или 2cosx+√3=0
x=π/2+πn, n∈Z
cosx=√3/2
x=+-π/6+2πn, n∈Z
2 votes
Thanks 7
furman1995
Спасибо тебе огромное))))
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Verified answer
2cos²x=√3sin(3π/2+x)2cos²x= -√3cosx
2cos²x+√3cosx=0
cosx(2cosx+√3)=0
cosx=0 или 2cosx+√3=0
x=π/2+πn, n∈Z cosx=√3/2
x=+-π/6+2πn, n∈Z