1)sin^4x-cos^4x=-sin4x 2) 4cos^2x+sinxcosx+3sin^2x-3=0 - вопрос №412851514442423230 от texin508 25.11.2021 20:14

Question

Алгебра: 1)sin^4x-cos^4x=-sin4x 2) 4cos^2x+sinxcosx+3sin^2x-3=0 решить уравнения

texin508 · Answer

1)sin⁴x-cos⁴x=-sin4x;⇒(sin²x+cos²x)(sin²x-cos²x)=-sin4x;⇒
⇒1·(-cos2x)=-2sin2x·cos2x;⇒cos2x=2sin2xcos2x;⇒
cos2x(1-2sin2x)=0;
cos2x=0;⇒2x=π/2+kπ;k∈Z;⇒x=π/4+kπ/2;k∈Z;
1-2sin2x=0;⇒sin2x=1/2;⇒2x=(-1)^k·π/6+kπ;⇒x=(-1)^k·π/12+kπ/2;
2)4cos²x+sinxcosx+3sin²x-3=0⇒3cos²x+cos²x+sinxcosx+3sin²x-3=0⇒
3(cos²x+sin²x)+cos²x+sinxcosx-3=0;⇒3+cos²x+sinxcosx-3=0;⇒
cosx(cosx+sinx)=0;
cosx=0;⇒x=π/2+kπ;k∈Z;
cosx+sinx=0;⇒cosx≠0;⇒cosx/cosx+sinx/cosx=0;⇒
1+tgx=0;⇒tgx=-1;⇒x₁=-π/4+2kπ;k∈Z.x₂=3π/4+2kπ.