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2001eka
@2001eka
July 2022
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Помогите, пожалуйста, срочно: [tex]y= \sqrt{sin^{2} -cos^{2} } [/tex] Найти D(y) область определения:
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армения20171
Sin²x-cos²x>=0
-(cos²x-sin²x)>=0
cos2x<=0
π/2+2πk<=2x<=3π/2+2πk
π/4+πk<=x<=3π/4+πk ;k€Z
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Answers & Comments
-(cos²x-sin²x)>=0
cos2x<=0
π/2+2πk<=2x<=3π/2+2πk
π/4+πk<=x<=3π/4+πk ;k€Z