Автор - Vov2998

Помогииииитеее
6cos2x + 8 sin^2x - 5 = 0
4 cos 2x + 10 cos x + 7 =0
3 sin 2x + 56 sin ^2x - 20 =0

Ответ

Автор - mymurkin
6(1-2sin²x)+8sinx-5=0
-12sin²x+8sinx+1=0
12sin²x-8sinx-1=0
sinx=t   |t|≤1
12t²-8t-1=0
D=16+12=28
t=4+√28/12=4+2√7/12=2+√7/6
t=4-√28/12=2-√7/6
sinx=2-√7/6
x=(-1)^narcsin(2-√7/6)+πn
sinx=2+√7/6
x=(-1)^narcsin(2+√7/6)+πn

4(2cos²x-1)+10cosx+7=0
8cos²x-4+10cosx+7=0
8cos²x+10cosx+3=0
cosx=t  |t|≤0
8t²+10t+3=0
D=25-24=1
t1=-5+1/8=-1/2
t2=-5-1/8=-3/4
cosx=-1/2
x=+-2π/3+2πn
cosx=-3/4
x=+-arccos(-3/4)+2πn

3sin2x+56sin²x-20=0
6sinx*cosx+56sin²x-20sin²x-20cos²x=0
36sin²x+6sinx*cosx-20cos²x=0
поделим на cos²x≠0
36tg²x+6tgx-20=0
tgx=t
18t²+3t-10=0
D=9+4*10*18=729=27²
t1=-3-27/18=-30/18=-5/3
t2=-3+27/18=24/18=4/3
tgx=-5/3
x=-arctg5/3+πn
tgx=4/3
x=arctg4/3+πn

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