Question

Помогите решить 140,141

Answer 1

$140.quad 1+2cos^2x+2sqrt2cdot sinx+cos2x=0\\star ; ; cos^2x=1-sin^2x; ; star \\star ; ; cos2x=cos^2x-sin^2x=(1-sin^2x)-sin^2x=1-2sin^2x; ; star \\\1+2-2sin^2x+2sqrt2cdot sinx+1-2sin^2x=0\\-4sin^2x+2sqrt2cdot sinx+4=0; |:(-2)\\2sin^2x-sqrt2cdot sinx-2=0\\D=2+4cdot 2cdot 2=18; ,\\a); ; (sinx)_1= frac{sqrt2-sqrt{18}}{4}= frac{sqrt2-3sqrt2}{4}=frac{-2sqrt2}{4}=-frac{sqrt2}{2}approx -0,71$

$x=(-1)^{n}arcsin(-frac{sqrt2}{2})+pi n=(-1)^{n+1}cdot frac{pi}{4}+pi n,; nin Z$

$b); ; (sinx)_2= frac{sqrt2+sqrt{18}}{4} = frac{4sqrt2}{4}=sqrt2approx 1,41 textgreater 1\\net; ; reshenij\\Otvet:; ; x=(-1)^{n+1}cdot frac{pi}{4}+pi n,; nin Z$

$141.quad cos(2x+ frac{2pi }{3})+4sin(x+frac{pi }{3})=2,5\\star ; ; 2x+ frac{2pi }{3} =2cdot (x+frac{pi}{3}); to ; alpha =x+frac{pi}{3}; ,; ; 2 alpha =2x+frac{2pi}{3}; star \\star ; ; cos2 alpha =1-2sin^2 alpha ; ; star \\\cos2 alpha +4sin alpha =2,5\\1-2sin^2 alpha +4sin alpha -2,5=0; |cdot (-1)\\2sin^2 alpha -4sin alpha +1,5=0; |cdot 2\\4sin^2 alpha -8sin alpha +3=0\\D/4=4^2-4cdot 3=4; ,\\a); ; (sin alpha )_1= frac{4-2}{4}=frac{1}{2}$

$alpha =(-1)^{n}cdot frac{pi}{6}+pi n,; nin Z\\x+frac{pi}{3}=(-1)^{n}cdot frac{pi}{6}+pi n,; nin Z\\x=-frac{pi}{3}+(-1)^{n}cdot frac{pi}{6}+pi n,; nin Z\\b); ; ; (sin alpha )_2=frac{4+2}{4}= frac{3}{2} textgreater 1; ; to ; ; net; reshenij \\Otvet :; ; x=-frac{pi}{3}+(-1)^{n}cdot frac{pi}{6}+pi n,; nin Z$

$P.S.; ; ax^2+2px+q=0; ,quad (2p, vdots , 2)\\D/4=p^2-aq ; ,; ; x_{1,2}= frac{-ppm sqrt{D/4}}{a}$