Решить. за . 1)2sin (40°+x)sin(x-50°)+1=0 2)4cos^2x+sinxcosx+3sin^2x-3=0 - вопрос №77785081240501024232303 от Vika2002588 06.07.2020 05:01

Question

Алгебра: Решить. за . 1)2sin (40°+x)sin(x-50°)+1=0 2)4cos^2x+sinxcosx+3sin^2x-3=0 3)cos^2x-3sinxsosx=-1

Vika2002588 · Answer

12sin (40°+x)sin(x-50°)+1=0 sin (40°+x)sin(x-50°)=-1/21/2*[cos(40+x-x+50)-cos(40+x+x-50)]=-1/2cos90-cos(2x-10)=-1/2cos(2x-10)=1/22x-10=-60+360k U 2x-10=60+360k2x=-50+360k U 2x=70+360kx=25+180k U x=35+180k,k∈z24cos^2x+sinxcosx+3sin^2x-3=0 4cos²x+sinxcosx+3sin²x-3cos²x-3sin²x=0cos²x+sinxcosx=0cosx(cosx+sinx)=0cosx=0⇒x=π/2+πk,k∈zcosx+sinx=0/cosx1+tgx=0tgx=-1⇒x=-π/4+πk,k∈z3cos^2x-3sinxcosx=-1cos²x-3sinxcosx+sin²x+cos²x=0 /cos²xtg²x+3tgx+2=0tgx=aa²+3a+2=0D=9-8=1 0a1+a2=-3 U a1+a2=2a1=-2⇒tgx=-2⇒x=-arctg2+πk,k∈za2=-1⇒tgx=-1⇒x=-π/4+πk,k∈z...