Question

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Answer 1

Ответ:

$1)\ \ sin^22x+sin^2x=\dfrac{9}{16}\\\\4sin^2x\cdot cos^2x+sin^2x=\dfrac{9}{16}\ \ ,\ \ \ 4sin^2x\cdot (1-sin^2x)+sin^2x=\dfrac{9}{16}\ \ ,\\\\4sin^2x-4sin^4x+sin^2x=\dfrac{9}{16}\ \ ,\ \ \ 4sin^4x-5sin^2x+\dfrac{9}{16}=0\ \ ,\\\\t=sin^2x\ ,\ \ t\in [\ 0\ ;\ 1\ ]\ :\ \ 4t^2-5t+\dfrac{9}{16}=0\ \ ,\ \ 64t^2-80t+9=0\ \ ,\\\\D/4=1600-576=1024=32^2\ \ ,\ \ t_1=\dfrac{8}{64}=\dfrac{1}{8}\ ,\ t_2=\dfrac{72}{64}=\dfrac{9}{8}>1$

$a)\ \ sin^2x=\dfrac{1}{8}\ \ ,\ \ \ sinx=\pm \dfrac{1}{2\sqrt2}\ \ ,\\\\x_1=(-1)^{n}\cdot arcsin\dfrac{1}{2\sqrt2}+\pi n\ ,\ n\in Z\\\\x_2=(-1)^{n+1}\cdot arcsin}\dfrac{1}{2\sqrt2}+\pi k\ ,\ k\in Z\\\\b)\ \ sin^2x=\dfrac{9}{8}>1\ \ \to \ \ \ x\in \varnothing \\\\Otvet:\ \ x_1=(-1)^{n}\cdot arcsin\dfrac{1}{2\sqrt2}+\pi n\ ,\\\\{}\qquad \qquad x_2=(-1)^{n+1}\cdot arcsin}\dfrac{1}{2\sqrt2}+\pi k\ ,\ n,k\in Z\ .$  

$2)\ \ \sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}>1\\\\\sqrt{(\sqrt{x-1}-2)^2}+\sqrt{(\sqrt{x-1}-3)^2}>1\\\\|\sqrt{x-1}-2|+|\sqrt{x-1}-3|>1\\\\t=\sqrt{x-1}\geq 0\ \ ,\ \ \ |\, t-2\, |+|\, t-3\, |>1\ \ ,\\\\a)\ t\leq 2:\ \ |t-2|+|t-3|=-t+2-t+3=-2t+5\ ,\ \ -2t+5>1\ ,\\\\2t<4\ \ ,\ \ t<2\ ,\\\\0\leq \sqrt{x-1}<2\ \ ,\ \ 0\leq x-1<4\ \ ,\ \ \underline{\ 1\leq x<5\ }\\\\b)\ \ 2<t\leq 3\ :\ \ |t-2|+|t-3|=t-2-t+3=1\notin (\, 2\, ;\, 3\, ]\\\\c)\ \ t>3\ :\ \ |t-2|+|t-3|=t-2+t-3=2t-5\ \ ,\ \ 2t-5>1\ \ ,$

$2t>6\ \ ,\ \ t>3\ \ ,\\\\\sqrt{x-1}>3\ \ ,\ \ x-1>9\ \ ,\ \ \underline{\ x>10\ }\\\\Otvet:\ \ x\in [\ 1\ ;\ 5\ )\cup (\ 10\ ;+\infty \, )\ .$