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: cosphi=frac{a}{sqrt{a^2+b^2}}; sinphi=frac{b}{sqrt{a^2+b^2}}$

В данном уравнении:

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

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