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maslovae420
@maslovae420
August 2022
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СРОЧНО ПОМОГИТЕ ПОЖАЛУЙСТА РЕШИТЬ)
Решите уравнения:
а) -2sin(x)+ √3=0
б) cos(3x+pi/3)-1=0
в) -2cos^2 (x)-5sin(x)-1=0
г) sin^2 (x)+4sin(x)cos(x)-5cos^2 (x)=0
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sedinalana
Verified answer
1
2sinx=√3
sinx=√3/2
x=(-1)^n*π/3+πn,n∈z
2
cos(3x+π/3)=1
3x+π/3=2πn
3x=-π/3=2πn
x=-π/9+2πn/3,n∈z
3
-2(1-sin²x)-5sinx-1=0
-2+2sin²x-5sinx-1=0
sinx=a
2a²-5a-3=0
D=25+24=49
a1=(5-7)/4=-1/2⇒sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn
a2=(5+7)/4=3⇒sinx=3>1 нет решения
4
Разделим на cos²x
tg²x+4tgx-5=0
tgx=a
a²+4a-5=0
a1+a2=-4 U a1*a2=-5
a1=-5⇒tgx=-5⇒x=-arctg5+πn,n∈z
a2=1⇒tgx=1⇒x=π/4+πn,n∈z
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Answers & Comments
Verified answer
12sinx=√3
sinx=√3/2
x=(-1)^n*π/3+πn,n∈z
2
cos(3x+π/3)=1
3x+π/3=2πn
3x=-π/3=2πn
x=-π/9+2πn/3,n∈z
3
-2(1-sin²x)-5sinx-1=0
-2+2sin²x-5sinx-1=0
sinx=a
2a²-5a-3=0
D=25+24=49
a1=(5-7)/4=-1/2⇒sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn
a2=(5+7)/4=3⇒sinx=3>1 нет решения
4
Разделим на cos²x
tg²x+4tgx-5=0
tgx=a
a²+4a-5=0
a1+a2=-4 U a1*a2=-5
a1=-5⇒tgx=-5⇒x=-arctg5+πn,n∈z
a2=1⇒tgx=1⇒x=π/4+πn,n∈z