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tolia12345
@tolia12345
August 2022
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помогите пожалуйста сделать))нужно с решением,
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hlopushinairina
A) sinx=1/2; ⇒x=π/6+kπ;k∈Z;
б) соsx/2=-1;⇒x/2=3π/2+2kπ;k∈Z;⇒x=3π+4kπ;k∈Z;x=π+2kπ;
в) ctg(2x+π/4)=-1;(2x+π/4)=3π/4+kπ;k∈Z;2x=3π/4-π/4+kπ=π/2+kπ;k∈Z;
г)2cos²x=1+sinx;⇒2(1-sin²x)=1+sinx;⇒sin²x+sinx-1=0;
sinx=y;⇒y²+y-1=0⇒y₁₂=-1/2⁺₋√1/4+1;
y₁=-1/2+√5/2=(√5-1)/2=0.61; y₂=-1/2-√5/2=-(1+√5)/2=-1.62;
-1<sinx<1;sinx=0.61;⇒x=(-1)^k·arcsin0.61+2kπ;k∈Z;
д)2sin²x-5sinxcosx+3cos²x=0;⇒2sin²x/cos²x-5sinxcosx/cos²x+3cos²x/cos²x=0⇒
2tg²x-5tgx+3=0;⇒tgx=y⇒
2y²-5y+3=0;
y₁₂=(5⁺₋√25-24)/4;
y₁=(5+1)/4=3/2;y₂=(5-1)/4=1;
tgx=3/2;⇒x=arctg3/2+kπ;k∈Z;
tgx=1⇒x=π/4+kπ;k∈Z.
2 votes
Thanks 1
tolia12345
спасибо огромнейшее)))
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Answers & Comments
б) соsx/2=-1;⇒x/2=3π/2+2kπ;k∈Z;⇒x=3π+4kπ;k∈Z;x=π+2kπ;
в) ctg(2x+π/4)=-1;(2x+π/4)=3π/4+kπ;k∈Z;2x=3π/4-π/4+kπ=π/2+kπ;k∈Z;
г)2cos²x=1+sinx;⇒2(1-sin²x)=1+sinx;⇒sin²x+sinx-1=0;
sinx=y;⇒y²+y-1=0⇒y₁₂=-1/2⁺₋√1/4+1;
y₁=-1/2+√5/2=(√5-1)/2=0.61; y₂=-1/2-√5/2=-(1+√5)/2=-1.62;
-1<sinx<1;sinx=0.61;⇒x=(-1)^k·arcsin0.61+2kπ;k∈Z;
д)2sin²x-5sinxcosx+3cos²x=0;⇒2sin²x/cos²x-5sinxcosx/cos²x+3cos²x/cos²x=0⇒
2tg²x-5tgx+3=0;⇒tgx=y⇒
2y²-5y+3=0;
y₁₂=(5⁺₋√25-24)/4;
y₁=(5+1)/4=3/2;y₂=(5-1)/4=1;
tgx=3/2;⇒x=arctg3/2+kπ;k∈Z;
tgx=1⇒x=π/4+kπ;k∈Z.