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Elena319
@Elena319
April 2021
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Срочно, помогите пожалуйста!!!
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hlopushinairina
1.2sin²x-3sinx+1=0;
-1≤sinx≤1;sinx=y;⇒
2y²-3y+1=0;
y₁,₂=[3⁺₋√(9-8)]/4=(3⁺₋1)/4;
y₁=4/4=1;
sinx=1;x=π/2+2nπ;n∈Z;
y₂=2/4=1/2;
sinx=1/2;x=(-1)ⁿ·π/6+nπ;n∈Z;
2.Cos²x+6sinx-6=0;⇒
1-sin²x+6sinx-6=0;
sin²x-6sinx+5=0;⇒sinx=y;-1≤y≤1;
y²-6y+5=0;
y₁,₂=3⁺₋√(9-5)=3⁺₋2;
y₁=5⇒нет решения
y₂=1⇒sinx=1;x=π/2+nπ;n∈Z;
3. cos2x+8sinx=3;⇒1-2sin²x+8sinx-3=0;
2sin²x-8sinx+2=0;⇒sinx=y;-1≤y≤1;
2y²-8y+2=0;⇒
y²-4y+1=0;
y₁,₂=2⁺₋√(4-1)=2⁺₋√3;
y₁=2+√3>1⇒нет решения
y₂=2-√3=2-1.73=0.267
sinx=0.267;x=(-1)ⁿ·arcsin0.267+nπ;n∈Z;
4.√3·tg²3x-3·tg3x=0;⇒
√3·tg3x(tg3x-√3)=0;
tg3x=0;⇒3x=nπ;n∈Z;x=nπ/3;n∈Z;
tg3x-√3=0;⇒
tg3x=√3;3x=π/3+nπ;n∈Z;x=π/9+nπ/3;n∈Z;
5.2sin²2x=(cosx+sinx)²;⇒2sin²2x=cos²x+2sinxcosx+sin²x;
2sin²2x=1+sin2x;⇒
2sin²2x-sin2x-1=0;
sinx=y;-1≤y≤1;
2y²-y-1=0;
y₁,₂=[1⁺₋√(1+8)]/4=(1⁺₋3)/4;
y₁=4/4=1;sinx=1;x=π/2+2nπ;n∈Z;
y₂=-2/4=-1/2;sinx=-1/2;x=(-1)ⁿ·5π/6+nπ;n∈Z
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Answers & Comments
-1≤sinx≤1;sinx=y;⇒
2y²-3y+1=0;
y₁,₂=[3⁺₋√(9-8)]/4=(3⁺₋1)/4;
y₁=4/4=1;
sinx=1;x=π/2+2nπ;n∈Z;
y₂=2/4=1/2;
sinx=1/2;x=(-1)ⁿ·π/6+nπ;n∈Z;
2.Cos²x+6sinx-6=0;⇒
1-sin²x+6sinx-6=0;
sin²x-6sinx+5=0;⇒sinx=y;-1≤y≤1;
y²-6y+5=0;
y₁,₂=3⁺₋√(9-5)=3⁺₋2;
y₁=5⇒нет решения
y₂=1⇒sinx=1;x=π/2+nπ;n∈Z;
3. cos2x+8sinx=3;⇒1-2sin²x+8sinx-3=0;
2sin²x-8sinx+2=0;⇒sinx=y;-1≤y≤1;
2y²-8y+2=0;⇒
y²-4y+1=0;
y₁,₂=2⁺₋√(4-1)=2⁺₋√3;
y₁=2+√3>1⇒нет решения
y₂=2-√3=2-1.73=0.267
sinx=0.267;x=(-1)ⁿ·arcsin0.267+nπ;n∈Z;
4.√3·tg²3x-3·tg3x=0;⇒
√3·tg3x(tg3x-√3)=0;
tg3x=0;⇒3x=nπ;n∈Z;x=nπ/3;n∈Z;
tg3x-√3=0;⇒
tg3x=√3;3x=π/3+nπ;n∈Z;x=π/9+nπ/3;n∈Z;
5.2sin²2x=(cosx+sinx)²;⇒2sin²2x=cos²x+2sinxcosx+sin²x;
2sin²2x=1+sin2x;⇒
2sin²2x-sin2x-1=0;
sinx=y;-1≤y≤1;
2y²-y-1=0;
y₁,₂=[1⁺₋√(1+8)]/4=(1⁺₋3)/4;
y₁=4/4=1;sinx=1;x=π/2+2nπ;n∈Z;
y₂=-2/4=-1/2;sinx=-1/2;x=(-1)ⁿ·5π/6+nπ;n∈Z