Решите 1. 3sin2x+8cos²x=7 2. cosx-2/cos(x/2)=2 - вопрос №86828591328722223122 от arifnabiev83 27.11.2020 17:22

Question

Алгебра: Решите 1. 3sin2x+8cos²x=7 2. cosx-2/cos(x/2)=2 3. 1+sin2x ×cosx=sin2x+cosx

arifnabiev83 · Answer

1. 3sin2x+8cos²x=76sinxcosx+8cos²x-7sin²x-7cos²x=07sin²x-6sinxcosx-1=0/cos²x7tg²x-6tgx-1=0tgx=t7t²-6t-1=0D=36+28=64t1=(6-8)/14=-1/7⇒tgx=-1/7⇒x=-arctg1/7+πk,k∈zt2=(6+8)/14=1⇒tgx=1⇒x=π/4+πk,k∈z2. (cosx-2)/cos(x/2)=2cos(x/2)≠0⇒x/2≠π/2+πk⇒x≠π+2πk,k∈z2cos²(x/2)-1-3-2cos(x/2)=0cos(x/2)=t2t²-2t-3=0D=4+24=28t1=(2-2√7)/4=0,5-0,5√7⇒cos(x/2)=0,5-0,5√7x/2=+-arccos(0,5-0,5√7)+2πkx=+-2arccos(0,5-0,5√7)+2πk,k∈zt2=0,5+0,5√7⇒cos(x/2)=0,5+0,5√7 1 нет решения3. 1+sin2x ×cosx=sin2x+cosx(sin2xcosx-sin2x)+(1-cosx)=0sin2x(cosx-1)-(cosx-1)=0(cosx-1)(sin2x-1)=0cosx=1⇒x=2πk,k∈zsin2x=1⇒2x=π/2+2πk⇒x=π/4+πk,k∈z...