Question

срочно и подробно пожалуйста ​

Answer 1

$1)\ \ 4\, sin3x\cdot cos3x>\sqrt2\\\\\star \ \ 2\, sina\cdot cosa=sin2a\ \ \to \ \ \ sina\cdot cosa=\dfrac{1}{2}\, sin2a\ \ \star \\\\2\cdot sin6x>\sqrt2\\\\sin\, 6x>\dfrac{\sqrt2}{2}\ \ \ \Rightarrow \ \ \ \dfrac{\pi}{4}+2\pi n<6x<\dfrac{3\pi}{4}+2\pi n\ ,\ n\in Z\ ,\\\\\\\dfrac{\pi}{24}+\dfrac{\pi n}{3}<x<\dfrac{3\pi }{24}+\dfrac{\pi n}{3}\ \ ,\ \ n\in Z$

$2)\ \ \dfrac{sin3x-cos3x}{sin3x+cos3x}<0\\\\\star \ \ sina-cosa=sina-sin(\dfrac{\pi}{2}-a)=2\, sin\dfrac{a-\frac{\pi}{2}+a}{2}\cdot cos\dfrac{a+\frac{\pi}{2}-a}{2}=\\\\=2\, sin(a-\dfrac{\pi}{4})\cdot cos\dfrac{\pi}{4}=\sqrt2\cdot sin(a-\dfrac{\pi}{4})\ \ \star \\\\\star \ \ sina+cosa=sina+sin(\dfrac{\pi}{2}-a)=2\, sin\dfrac{\pi}{4}\cdot cos(a-\dfrac{\pi}{4})=\sqrt2\cdot cos(a-\dfrac{\pi}{4})\ \ \star$

$\dfrac{\sqrt2\cdot sin(3x-\frac{\pi}{4})}{\sqrt2\cdot cos(3x-\frac{\pi}{4})}<0\ \ \ \Rightarrow \ \ \ tg(3x-\dfrac{\pi}{4})<0\ \ \ \Rightarrow \\\\\\-\dfrac{\pi}{2}+\pi n<3x-\dfrac{\pi}{4}<\pi n\ ,\ n\in Z\\\\\\-\dfrac{\pi}{4}+\pi n<3x<\dfrac{\pi}{4}+\pi n\ ,\ n\in Z\\\\\\-\dfrac{\pi}{12}+\dfrac{\pi n}{3}<x<\dfrac{\pi}{12}+\dfrac{\pi n}{3}\ ,\ n\in Z$

$3)\ \ 3sin2x+8cos^2x\geq 7\\\\6\cdot sinx\cdot cosx+8cos^2x-7(sin^2x+cos^2x)\geq 0\\\\-7sin^2x+6\cdot sinx\cdot cosx+cos^2x\geq 0\ \Big|:cos^2x\ne 0\\\\-7tg^2x+6tgx+1\geq 0\ \ \ \Rightarrow \ \ \ 7tg^2x-6tgx-1\leq 0\ ,\\\\t=tgx\ \ ,\ \ \ 7t^2-6t-1\leq 0\ \ ,\ \ t_1=-\dfrac{1}{7}\ \ ,\ \ t_2=1\ ,\\\\7(t+\dfrac{1}{7})(t-1)\leq 0\ \ ,\ \ \ \ +++[-\dfrac{1}{7}\ ]---[\ 1\ ]+++$

$\-\dfrac{1}{7}\leq t\leq 1\ \ \ \Rightarrow \ \ \ -\dfrac{1}{7}\leq tgx\leq 1\ \ ,\ \ -arctg\dfrac{1}{7}+\pi n\leq x\leq \dfrac{\pi}{4}+\pi n\ ,\ n\in Z$