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Merguha
@Merguha
July 2022
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помогите cosx+cos2x=0
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yuraborovkov98
Cosx+ cos^2x-sin^2x=0
cos^2x+ cosx-1+cos^2x=0
2cos^2x+cosx-1=0, cosx=y,
2y^2+ y - 1 = 0, D=9, y1=1/2, y2=-1, Значит cosx=1/2, x1=+(-)п/3+2Пк, к - целое число и cosx=-1, x2=п+2пn, n - целое число.
Ответ: x1=+(-)п/3+2Пк и x2=п+2пn
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Answers & Comments
cos^2x+ cosx-1+cos^2x=0
2cos^2x+cosx-1=0, cosx=y,
2y^2+ y - 1 = 0, D=9, y1=1/2, y2=-1, Значит cosx=1/2, x1=+(-)п/3+2Пк, к - целое число и cosx=-1, x2=п+2пn, n - целое число.
Ответ: x1=+(-)п/3+2Пк и x2=п+2пn