Question

ПОМОГИТЕ ПОЖАЛУЙСТА СРОЧНО ​

Answer 1

$1)\ \ cos^2x>\dfrac{1}{4}\ \ \to \ \ \ \dfrac{1+cos2x}{2}>\dfrac{1}{4}\ \ ,\ \ 1+cos2x>\dfrac{1}{2}\ \ ,\ \ \, cos2x>-\dfrac{1}{2}\\\\\\\dfrac{2\pi}{3}+2\pi n<2x<\dfrac{4\pi}{3}+2\pi n\ ,\ n\in Z\\\\\\\dfrac{\pi}{3}+\pi n<x<\dfrac{2\pi}{3}+\pi n\ ,\ n\in Z\\\\\\x\in \Big(\ \dfrac{\pi}{3}+\pi n\ ;\ \dfrac{2\pi}{3}+\pi n\ \Big)\ ,\ n\in Z$

$2)\ \ sin^2x<\dfrac{1}{4}\ \ \to \ \ \ \dfrac{1-cos2x}{2}<\dfrac{1}{4}\ \ ,\ \ 1-cos2x<\dfrac{1}{2}\ \ ,\ \ \, cos2x>\dfrac{1}{2}\\\\\\-\dfrac{\pi}{3}+2\pi n<2x<\dfrac{\pi}{3}+2\pi n\ ,\ n\in Z\\\\\\-\dfrac{\pi}{6}+\pi n<x<\dfrac{\pi}{6}+\pi n\ ,\ n\in Z\\\\\\x\in \Big(\, -\dfrac{\pi}{6}+\pi n\ ;\ \dfrac{\pi}{6}+\pi n\ \Big)\ ,\ n\in Z$

$3)\ \ 2sin^2x+\sqrt3sinx-3>0\\\\t=sinx\ \ ,\ \ |\, t\, |\leq 1\ \ ,\ \ \ 2t^2+\sqrt3t-3>0\ \ ,\ \ D=3+24=27\ \ ,\\\\t_{1}=\dfrac{-\sqrt3-3\sqrt3}{4}=\dfrac{-4\sqrt3}{4}=-\sqrt3\approx -1,73<-1\\\\t_2=\dfrac{-\sqrt3+3\sqrt3}{4}=\dfrac{2\sqrt3}{4}=\dfrac{\sqrt3}{2}\approx 0,87\\\\\\2\, (t+\sqrt3)(t-\frac{\sqrt3}{2})>0\ \ ,\ \ \ \ +++(-\sqrt3)---(\frac{\sqrt3}{2})+++\\\\t\in (-\infty ;-\sqrt3)\cup (\frac{\sqrt3}{2}\ ;+\infty )$

$\left\{\begin{array}{l}t\in (-\infty ;-\sqrt3)\cup (\frac{\sqrt3}{2}\ ;+\infty )\\t\in [-1\, ;\ 1\ ]\end{array}\right\ \ \ \Rightarrow \ \ \ t\in (\frac{\sqrt3}{2}\ ;\ 1\ ]\\\\\\\dfrac{\sqrt3}{2}<sinx\leq 1\ \ ,\ \ \left\{\begin{array}{l}sinx>\dfrac{\sqrt3}{2}\\sinx\leq 1\end{array}\right\ \ \ \Rightarrow \ \ \ x\in \Big(\ \dfrac{\pi}{3}+2\pi n\ ;\ \dfrac{2\pi }{3}+2\pi n\Big)\ ,\ n\in Z$

$4)\ \ 2\, sin^2x-3\, sinx+1\geq 0\\\\t=sinx\ \ ,\ \ |\, t\, |\leq 1\ \ ,\ \ \ 2t^2-3t+1\geq 0\ \ ,\ \ \ D=9-8=1\ \ ,\\\\t_1=\dfrac{1}{2}\ ,\ \ t_2=1\ \ ,\ \ \ 2(t-0,5)(t-1)\geq 0\\\\znaki:\ \ \ +++(0,5)---(1)+++\ \ ,\ \ t\in (-\infty \, ;\ 0,5\ ]\cup [\ 1\ ;+\infty \, )\\\\\\\left\{\begin{array}{l} t\in (-\infty \, ;\ 0,5\ ]\cup [\ 1\ ;+\infty \, )\\t\in [-1\ ;\ 1\ ]\end{array}\right\ \ \ \Rightarrow \ \ t\in [-1\, ;\ 0,5\ ]\cup \{1\}$

$\left[\begin{array}{l}-1\leq sinx\leq 0,5\\sinx=1\end{array}\right\ \ \left[\begin{array}{l}-\dfrac{7\pi}{6}+2\pi n\leq x\leq \dfrac{\pi}{6}+2\pi n\ ,\ n\in Z\\x=\dfrac{\pi}{2}+2\pi n\ ,\ n\in Z\end{array}\right\\\\\\x\in \Big(-\dfrac{7\pi}{6}+2\pi n\ ;\ \dfrac{\pi}{6}+2\pi n\ \Big)\cup \Big\{\dfrac{\pi}{2}+2\pi n\Big\}\ ,\ n\in Z$