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katyamatveeva
@katyamatveeva
July 2022
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даю 99 балов
Один пример на фото,тема:решение тригонометрических уравнений
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sedinalana
Verified answer
2*2sin(x/2)cos(x/2)+cos²(x/2)-sin²(x/2)-2*(sin²(x/2)+cos²(x/2))=0
4sin(x/2)cos(x/2)+cos²(x/2)-sin²(x/2)-2sin²(x/2)-2cos²(x/2)=0
3sin²(x/2)-4sin(x/2)cos(x/2)+cos²(x/2)=0 /cos²(x/2)
3tg²(x/2)-4tg(x/2)+1=0
tg(x/2)=a
3a²-4a+1=0
D=16-12=4
a1=(4-2)/6=1/3⇒tg(x/2)=1/3⇒x/2=arctg1/3+πn⇒x=2arctg1/3+2πn,n∈z
a2=(4+2)/6=1⇒tg(x/2)=1⇒x/2=π/4+πk⇒x=π/2+2πk,k∈z
3 votes
Thanks 3
katyamatveeva
ОГРОМНЕЙШЕЕ СПАСИБОООО!!!!!!!!!!!!!
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Answers & Comments
Verified answer
2*2sin(x/2)cos(x/2)+cos²(x/2)-sin²(x/2)-2*(sin²(x/2)+cos²(x/2))=04sin(x/2)cos(x/2)+cos²(x/2)-sin²(x/2)-2sin²(x/2)-2cos²(x/2)=0
3sin²(x/2)-4sin(x/2)cos(x/2)+cos²(x/2)=0 /cos²(x/2)
3tg²(x/2)-4tg(x/2)+1=0
tg(x/2)=a
3a²-4a+1=0
D=16-12=4
a1=(4-2)/6=1/3⇒tg(x/2)=1/3⇒x/2=arctg1/3+πn⇒x=2arctg1/3+2πn,n∈z
a2=(4+2)/6=1⇒tg(x/2)=1⇒x/2=π/4+πk⇒x=π/2+2πk,k∈z