Question

ПОМОГИТЕ!!!
Решите тригонометрическое уравнение: 2 sin^2 x + 3sin 2x + 2=0

Answer 1

2sin^2x-2√3sinxcosx=0
sinx(sinx-√3cosx)=0
x=Пk ;  k∈z
tgx=√3
x=П/3+Пn; n∈Z

Answer 2

$2sin^2(x)+3sin(2x)+2=0\2sin^2(x)+6sin(x)cos(x)+2=0\2sin^2(x)+6sin(x)cos(x)+2sin^2(x)+2cos^2(x)=0\4sin^2(x)+6sin(x)cos(x)+2cos^2(x)=0\cos^2(x)neq 0\cos(x)neq 0\xneq frac{pi}{2} +pi k \4tg^2(x)+6tg(x)+2=0\tg(x)=t\4t^2+6t+2=0\a-b+c=0=>\x_1=-1\x_2=-frac{1}{2} \tg(x)=-1\x=arctg(-1)+pik\x=-frac{pi}{4} +pi k\tg(x)=-frac{1}{2} \x=-arctg(frac{1}{2})+pik$

k∈Z