Question

помогите!! с решением, хотя бы кратким ​

Answer 1

ОДЗ: х≥0 , x≠0    => x∈(0;+∞);

Условие: =

$\displaystyle \sqrt{12} cos\bigg(\frac{\pi}{8}\sqrt{x}\sqrt{\frac{4}{x}+3-x } \ \bigg)+\sqrt{12} sin\bigg(\frac{\pi}{8}\ x\sqrt{\frac{4}{x^2}+\frac{3}{x}-1 } \ \bigg)=2\sqrt{6}$

$\displaystyle \sqrt{12} cos\bigg(\frac{\pi}{8}\sqrt{x}\sqrt{\frac{4}{x}+3-x } \ \bigg)+\sqrt{12} sin\bigg(\frac{\pi}{8}\ x\sqrt{\frac{4}{x^2}+\frac{3}{x}-1 } \ \bigg)=2\sqrt{6} \\ \\ cos\bigg(\frac{\pi}{8}\sqrt{x}\sqrt{\frac{4}{x}+3-x } \ \bigg)+ sin\bigg(\frac{\pi}{8}\ x\sqrt{\frac{4}{x^2}+\frac{3}{x}-1 } \ \bigg)=\frac{2}{\sqrt{2}}$

$\displaystyle cos\bigg(\frac{\pi}{8}\sqrt{x}\sqrt{\frac{4}{x}+3-x } \ \bigg)+ sin\bigg(\frac{\pi}{8}\ \boldsymbol{\sqrt{x}\sqrt{x}}\sqrt{\frac{4}{x^2}+\frac{3}{x}-1 } \ \bigg)=\frac{2}{\sqrt{2}}\\ \\ \displaystyle cos\bigg(\frac{\pi}{8}\sqrt{x}\sqrt{\frac{4}{x}+3-x } \ \bigg)+ sin\bigg(\frac{\pi}{8}\ \boldsymbol{\sqrt{x}}\sqrt{\frac{4}{\boldsymbol{x}}+\boldsymbol{3}-\boldsymbol{x} } \ \bigg)=\frac{2}{\sqrt{2}}$

$\displaystyle \frac{\pi}{8}\sqrt{x} \sqrt{\frac{4}{x}+3-x } = t \\\\ \frac{\pi}{8}\sqrt{x} \sqrt{\frac{4+3x-x^2}{x} } = t\\\\\frac{\pi}{8} \sqrt{4+3x-x^2} = t$

$\displaystyle cos(t) + sin(t) = \sqrt{2}\\\\ \sqrt{2}\bigg(\frac{1}{\sqrt{2}} cos(t) +\frac{1}{\sqrt{2}} sin(t)\bigg) = \sqrt{2} \quad\quad\quad\bigg|\frac{1}{\sqrt{2}}=cos\bigg(\frac{\pi}{4}\bigg)=sin\bigg(\frac{\pi}{4}\bigg)\bigg|\\\\\\cos\bigg(\frac{\pi}{4}\bigg)cos(t) + sin\bigg(\frac{\pi}{4}\bigg)sin(t) =1 \\\\~\bigg|cos(\alpha-\beta) = cos(\alpha)*cos(\beta)+sin(\alpha)*sin(\beta) \bigg|\\\\\\cos\bigg(\frac{\pi}{4}-t\bigg) = 1 \\\\\\\frac{\pi}{4}-t = 2\pi n , n\in Z$

$\displaystyle t=\frac{\pi}{4}-2\pi n , n \in Z \\\\ \pi\sqrt{4+3x-x^2} = 2\pi - 16\pi n , n\in Z \\\\\sqrt{4+3x-x^2} = 2 - 16 n , n\in Z$

$\sqrt{a}\geq 0~~ => n\leq 0 , n\in Z$

$\displaystyle 4+3x-x^2 = (2 - 16 n )^2\\x^2-3x-4+(4-64n+256n^2)=0\\D = 9 +4(-256n^2+64n) = -(32n+9)(32n-1)\\D<0 \quad\quad\bigg| n\in \bigg(-\infty;-\frac{9}{32}\bigg)\bigg\cup\bigg(\frac{1}{32};+\infty\bigg)\bigg| \\\\ D\geq 0\quad\quad\bigg| n\in \bigg[-\frac{9}{32};\frac{1}{32}\bigg]\bigg|$

Решение есть только если n=0;

$\displaystyle n=0\\x(x-3) = 0\\x_1 = 0,~x_2 = 3\\ x=0 \notin \text{ODZ}$

Ответ: х=3

Answer 2

Ответ:

$\displaystyle \sqrt{12}\, cos\Big(\frac{\pi }{8}\cdot \sqrt{x}\cdot \sqrt{\frac{4}{x}+3-x}\Big)+\sqrt{12}sin\Big(\frac{\pi }{8}\cdot \sqrt{x}\cdot \sqrt{\frac{4}{x^2}+\frac{3}{x}-1}\Big)=2\sqrt6\ \Big|:\sqrt{12}\\\\\\cos\Big(\frac{\pi }{8}\cdot \sqrt{x}\cdot \sqrt{\frac{4+3x-x^2}{x}}\Big)+sin\Big(\frac{\pi }{8}\cdot x\cdot \sqrt{\frac{4+3x-x^2}{x^2}}\Big)=\frac{\sqrt{24}}{\sqrt{12}}$

$\displaystyle cos\Big(\frac{\pi }{8}\cdot \sqrt{x}\cdot \frac{\sqrt{4+3x-x^2}}{\sqrt{x}}\Big)+sin\Big(\frac{\pi }{8}\cdot x\cdot \frac{\sqrt{4+3x-x^2} }{x}\Big)=\sqrt{2}\\\\\\ODZ:\ x>0\ \ \ \ i\ \ \ \ 4+3x-x^2\geq 0\ \ ,\\\\4+3x-x^2=-(x^2-3x-4)=-(x+1)(x-4)\geq 0\ \ ,\ \ x\in [-1\ ;\ 4\ ]\\\\x>0\ \ \ i\ \ \ x\in [-1\ ;\ 4\ ]\ \ \ \Rightarrow \ \ \underline {\ x\in (\ 0\ ;\ 4\ ]\ }\\\\\\cos\Big(\frac{\pi }{8}\cdot \sqrt{4+3x-x^2}\Big)+sin\Big(\frac{\pi }{8}\cdot \sqrt{4+3x-x^2}\Big)=\sqrt{2}$

$\displaystyle t=\frac{\pi }{8}\cdot \sqrt{4+3x-x^2}\ ,\ \ t\geq 0\ \Rightarrow \ \ \ \ cost+sint=\sqrt2\ \ ,\\\\\\\sqrt2\cdot \Big(\frac{1}{\sqrt2}\cdot cost+\frac{1}{\sqrt2}\cdot sint\Big)=\sqrt2\ \ ,\ \ \ \frac{1}{\sqrt2}=cos\frac{\pi}{4}=sin\frac{\pi}{4}\ \ ,\\\\\\sin\frac{\pi}{4}\cdot cost+cos\frac{\pi}{4}\cdot sint=1\ \ ,\\\\\\sin\Big(\frac{\pi}{4} +t\Big)=1\ \ ,\ \ \frac{\pi}{4} +t=\frac{\pi}{2}+2\pi n\ ,\ \ t=\frac{\pi}{4}+2\pi n\ ,\ n\in Z$

$\displaystyle \frac{\pi }{8}\cdot \sqrt{4+3x-x^2}=\frac{\pi }{4}+2\pi n\ ,\ n\in Z\\\\\frac{1}{8}\cdot \sqrt{4+3x-x^2}=\frac{1}{4}+2n\ ,\ n\in Z\\\\\sqrt{4+3x-x^2}=2+16n\ ,\ \ 2+16n\geq 0\ ,\ n\in Z\\\\4+3x-x^2=4+64n+256n^2\ \ ,\ \ n\geq -0,125,\ n\in Z\\\\x^2-3x+64n+256n^2=0\ \ ,\ n=0,1,2,\, ...$

$D=9-4(64n+256n^2)=9-256n-1024n^2=-(1024n^2+256n-9)\\\\D=9\ ,\ \ pri\ \ n=0\\\\1024n^2+256n-9=0\ ,\ D/4=25600=160^2\ \ ,\\\\n_1=-\frac{288}{1024}=-\frac{9}{32}\notin Z\ \ ,\ \ n_2=\frac{32}{1024}=\frac{1}{32}\notin Z\ ,$

$D=-(1024n^2+256n-9)=-(32n+9)(32n-1)\\\\D<0\ \ pri\ \ n\in (-\infty;-\frac{9}{32}\, )\cup (\frac{1}{32}\, ;+\infty )\ \ \to \ \ x\in \varnothing \\\\D>0\ \ pri\ \ n\in (-\frac{9}{32}\, ;\, \frac{1}{32}\, )\ \ \to \ \ n\notin Z\ ,\ krome\ n=0\ ,\\\\n=0:\ \ x^2-3x=0\ \ ,\ \ x(x-3)=0\ \ ,\ \ x_1=0\notin ODZ\ ,\ \ \ \underline {\ x_2=3\ }\\\\Otvet:\ x=3\ .$