Question

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

Answer 1

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

k∈Z

Answer 2

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