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drwnd
@drwnd
July 2022
1
39
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sin2x-cos2x= \sqrt{2} sin3x
приводя левую часть к виду sqrt{2}sin (pi\4-2x) и деля обе части на sqrt{2} получаем
sin (pi\4-2x)=sin3x
подскажите, что делать дальше? пожалуйста
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sedinalana
Verified answer
Sina-sinb=2sin[(a-b)/2]*cos[(a+b)/2]
---------------------------------------------------------
sin2x-sin(π/2-2x)=√2sin3x
2sin(2x-π/4)*cosπ/4=√2*sin3x
2*√2/2*sin(2x-π/4)=√2*sin3x
√2*sin(2x-π/4)=√2*sin3x
sin(2x-π/4)=sin3x
sin(2x-π/4)-sin3x=0
2sin(-x/2-π/8)*cos5x/2-π/8)=0
-2sin(x/2+π/8)*cos(5x/2-π/8)=0
sin(x/2+π/8)=0⇒x/2+π/8=πk⇒x/2=-π/8+πk⇒x=-π/4+2πk,k∈z
cos(5x/2-π/8)=0⇒5x/2-π/8=π/2+πk⇒5x/2=5π/8+πk⇒x=π/4+2πk/5,k∈z
2 votes
Thanks 3
drwnd
большое спасибо
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Answers & Comments
Verified answer
Sina-sinb=2sin[(a-b)/2]*cos[(a+b)/2]---------------------------------------------------------
sin2x-sin(π/2-2x)=√2sin3x
2sin(2x-π/4)*cosπ/4=√2*sin3x
2*√2/2*sin(2x-π/4)=√2*sin3x
√2*sin(2x-π/4)=√2*sin3x
sin(2x-π/4)=sin3x
sin(2x-π/4)-sin3x=0
2sin(-x/2-π/8)*cos5x/2-π/8)=0
-2sin(x/2+π/8)*cos(5x/2-π/8)=0
sin(x/2+π/8)=0⇒x/2+π/8=πk⇒x/2=-π/8+πk⇒x=-π/4+2πk,k∈z
cos(5x/2-π/8)=0⇒5x/2-π/8=π/2+πk⇒5x/2=5π/8+πk⇒x=π/4+2πk/5,k∈z