2cos^2x/2-3 sin(п-x/2)cos(2п-x/2)+7sin^2x/2=3 - вопрос №372364822232227223 от ibragimgadjizad 26.04.2020 21:42

Question

Алгебра: 2cos^2x/2-3 sin(п-x/2)cos(2п-x/2)+7sin^2x/2=3

ibragimgadjizad · Answer

2cos²x/2-3sinx/2*cosx/2+7sin²x/2-3sin²x/2-3cos²x/2=0
4sin²x/2-3sinx/2*cosx/2-cos²x/2=0/cos²x/2≠0
4tg²x/2-3tgx/2-1=0
tgx/2=a
4a²-3a-1=0
D=9+16=25
a1=(3-5)/8=-1/4⇒tgx/2=-1/4⇒x/2=-arctg1/4+πn⇒x=-2arctg1/4+2πn
a2=(3+5)/8=1⇒tgx/2=1⇒x/2=π/4+πn⇒x=π/2+2πn