Question

Помогите пожалуйста с решением уравнений

Answer 1

${bf 1)sqrt{(x+6)(x+1)}=6} \ \ (x+6)(x+1)=36 \ \ x^2+x+6x+6=36 \ \ x^2+7x-30=0 \ \ D = 49 + 120 = 169\ \ x_{12}=frac{-7 pm 13}{2}= left { {{3} atop {-10}} right$ 

${bf 2)log_{27}x+log_{9}x+log_{3}x=11} \ \ log_{3^3}x+log_{3^2}x+log_{3}x=11 \ \ frac{1}{3}log_{3}x+frac{1}{2}log_{3}x+log_{3}x=11 \ \ (frac{1}{3}+frac{1}{2}+1)log_{3}x=11 \ \ (frac{2}{6}+frac{3}{6}+frac{6}{6})log_{3}x=11 \ \ frac{11}{6}log_{3}x=11 \ \ log_{3}x=frac{11 cdot 6}{11} = 6 \ \ x = 3^6 = 729$ 

${bf 3)sin(3x)+0.5=0} \ \ sin(3x)=-0.5 \ \ 3x= left { {{frac{7 pi}{6}+2 pi n} atop {frac{-pi}{6} + 2 pi n}} right \ \ x=left { {{frac{1}{18}(12 pi n+7 pi)} atop {frac{1}{18}(12 pi n - pi)}} right$

${bf 4)cos(sin{2x})+sin(sin{2x})=-1} | ^2\ \ cos^2(sin{2x})+sin^2(sin{2x})+2cos(sin{2x})sin(sin{2x})=1 \ \ 1 + 2cos(sin{2x})sin(sin{2x})=1 \ \ 2cos(alpha)sin(alpha) = sin(2alpha) \ \ sin(2 cdot sin(2x))=0 \ \ 2 cdot sin(2x) = pi n, n in mathbb {Z} \ \ sin(2x) = frac{pi n}{2}, n in mathbb {Z} \ \ 2x = arcsin{frac{pi n}{2}}, n in mathbb {Z} \ \ x = frac{1}{2}arcsin{frac{pi n}{2}}, n in mathbb {Z}$