Question

$11sin(2x)+6cos^2(x)+6=0$

Answer 1

$11sin(2x)+6cos^2x+6=0\\11sin(2x)+6*\frac{1+cos(2x)}{2} +6=0\\2x=a\\11sina+3+3cosa+6=0\\11sina+3cosa=-9$

используем метод вспомогательного угла:

$asinx+bcosx=c\\\frac{a}{\sqrt{a^2+b^2}}*sinx+\frac{b}{\sqrt{a^2+b^2}}*cosx=\frac{c}{\sqrt{a^2+b^2}}\\(\frac{a}{\sqrt{a^2+b^2}})^2+(\frac{b}{\sqrt{a^2+b^2}})^2=1 \Rightarrow \exists \ \phi: \ cos\phi=\frac{a}{\sqrt{a^2+b^2}}; sin\phi=\frac{b}{\sqrt{a^2+b^2}}$

В данном уравнении:

$11sina+3cosa=-9\\\frac{11}{\sqrt{11^2+3^2}} *sina+\frac{3}{\sqrt{11^2+3^2}} *cosa=\frac{-9}{\sqrt{11^2+3^2}} \\cos\phi*sina+sin\phi*cosa=\frac{-9}{\sqrt{130}} \\sin(a+\phi)=\frac{-9}{\sqrt{130}} \\\phi=arcsin(\frac{3}{\sqrt{130}} )$

$sin(a+arcsin(\frac{3}{\sqrt{130}} ))=\frac{-9}{\sqrt{130}}\\a=(-1)^n*arcsin(\frac{-9}{\sqrt{130}})-arcsin(\frac{3}{\sqrt{130}} )+\pi n\\2x=(-1)^n*arcsin(\frac{-9}{\sqrt{130}})-arcsin(\frac{3}{\sqrt{130}} )+\pi n\\x=\frac{(-1)^n*arcsin(\frac{-9}{\sqrt{130}})}{2} -\frac{arcsin(\frac{3}{\sqrt{130}} )}{2} +\frac{\pi n}{2},\ n \in Z$