Question

Помогите 4,5,6
Во вложении
С объяснением, пожалуйста.

Answer 1

$1); ; cosx=frac{1}{sqrt{10}}\\frac{3pi }{2}leq xleq 2pi ; ; to ; ; tgx textless 0\\1+tg^2x=frac{1}{cos^2x}; ; to ; ; tgx=pm sqrt{frac{1}{cos^2x}-1}\\tgx textless 0; ; to ; ; tgx=-sqrt{frac{1}{1/10}-1}=-sqrt{9}=-3\\2); ; sinx=-frac{5}{sqrt{26}}\\pi leq xleq frac{3pi }{2}; ; to ; ; tgx textgreater 0\\tgx=pm sqrt{frac{1}{cos^2x}-1}=pm sqrt{frac{1}{1-sin^2x}-1}=pm sqrt{frac{1-1+sin^2x}{1-sin^2x}}=pm sqrt{frac{sin^2x}{1-sin^2x}}\\tgx textgreater 0; ; to ; ; tgx=+sqrt{frac{25/26}{1-25/26}}=sqrt{frac{25}{26-25}}=5$

$3); ; sinx=-frac{2sqrt2}{3}\\frac{3pi }{2}leq xleq 2pi ; ; to ; ; cosx textgreater 0\\sin^2x+cos^2x=1; ; to ; ; cos^2x=1-sin^2x; ,; cosx=pm sqrt{1-sin^2x}\\cosx textgreater 0; ; to ; ; cosx=+sqrt{1-(frac{2sqrt2}{3})^2}=sqrt{1-frac{8}{9}}=sqrt{frac{1}{9}}=frac{1}{3}\\3cosx=3cdot frac{1}{3}=1$

$4); ; cosx= frac{2sqrt6}{5}\\frac{3pi}{2}leq xleq 2pi ; ; to ; ; sinx textless 0\\sinx=-sqrt{1-cos^2x}=-sqrt{1-frac{24}{25}}=-frac{1}{5}\\5sinx=-5cdot frac{1}{5}=-1\\5); ; sinx=-0,2\\cos2x=cos^2x-sin^2x=(1-sin^2x)-sin^2x=1-2sin^2x=\\=1-2(-0,2)^2=1-2cdot 0,04=0,92\\6); ; sin3x=0,6\\frac{10sin6x}{3cos3x}=frac{10cdot 2sin3xcdot cos3x}{3cos3x}=frac{20}{3}cdot sin3x=frac{20}{3}cdot 0,6=4$