:1. 2sin^2x + 3cos^2x = 5sinx cosx 2. 2cos^2x + - вопрос №1223252221322694607 от BlazeySpace01 25.01.2020 05:38

Question

Алгебра: :1. 2sin^2x + 3cos^2x = 5sinx cosx 2. 2cos^2x + cosx = 1 3. 2cos^2 12x + 2 корень 2 cos 12x sin 12x + sin^2 12x = 0 4. 16sin x = cosx 5. cos^2x + 70sinx + 143 = 0 6. cos^2x + 23cosx = 24 7. cos^2x = 24cosx

BlazeySpace01 · Answer

2sin²x+3cos²x=5sinx*cosx |:cos²x2tg²-5tgx+3=0D=25-24=1tgx=1.5; x1=arctg1.5+πn,n€Ztgx=1; x2=-π/4+πn,n€Z2) 2cos²x+cosx=12cos²x+cosx-1=0пусть cosx=t(|t|≤1),имеем2t²+t-1=0t1=1/2; t2=-1заменacosx=1/2 и cosx=-1x1=±π/6+2πn; x2=2πn3) 2cos²12x+2√2cos12xsin12x+sin²12x=0tg²12x+2√2tgx+2=0(tg12x+√2)=0tg12x=-√2x=-1/12arctg√2+πn/124)16sinx=cosx |:cosx16tgx=1tgx=1/16x=arctg(1/16) + πnдалее аналогично, решать лень уже...