Question

Срочно даю 25 баллов

Answer 1

A

$cosx\geq \frac{\sqrt{2}}{2}$

$x_{1} =\frac{\pi }{4}+2\pi *n\\\\x_{2} =\frac{7\pi }{4}+2\pi *n$

$\frac{1}{4}(8\pi *n-\pi ) \leq x\leq \frac{1}{4}(8\pi *n+\pi )$ , n∈Z

$sinx\leq-\frac{\sqrt{2}}{2}$

$x_{1} =\frac{5\pi }{4}+2\pi *n\\\\x_{2} =\frac{7\pi }{4}+2\pi *n$

$\frac{1}{4}(8\pi *n-3\pi ) \leq x\leq \frac{1}{4}(8\pi *n-\pi )$ , n∈Z

$cosx<\frac{\sqrt{3}}{2}$

$x_{1} =\frac{\pi }{6}+2\pi *n\\\\x_{2} =\frac{11\pi }{6}+2\pi *n$

$\frac{1}{6}(12\pi *n+\pi )< x<\frac{1}{6}(12\pi *n+11\pi )$ , n∈Z

$tgx>\sqrt{3}}$

$x=\frac{\pi }{3}+\pi *n$

$pi *n-\frac{2\pi }{3}< x<pi *n-\frac{\pi }{2}$ , n∈Z

B

$2sinx+\sqrt{2}\geq 0,\\2sinx\geq -\sqrt{2},\\sinx\geq-\frac{\sqrt{2}}{2}$

$x_{1} =\frac{5\pi }{4}+2\pi *n\\\\x_{2} =\frac{7\pi }{4}+2\pi *n$

$\frac{1}{4}(8\pi *n-\pi ) \leq x\leq \frac{1}{4}(8\pi *n+5\pi )$ , n∈Z

$sin\frac{x}{2}\geq \frac{1}{2}$

$\frac{x_1}{2} =\frac{5\pi }{6}+2\pi *n\\\frac{x_2}{2} =\frac{\pi }{6}+2\pi *n\\\\x_1 =\frac{5\pi }{3}+4\pi *n\\x_2 =\frac{\pi }{3}+4\pi *n$

$\frac{1}{3}(12\pi *n+\pi ) \leq x\leq \frac{1}{3}(12\pi *n+5\pi )$ , n∈Z

или

$sin\frac{x}{2}\geq \frac{1}{2}$

$\\sin\frac{x}{2}=$±$\sqrt{\frac{1-cosx}{2}}$ ,

±$\sqrt{\frac{1-cosx}{2}}=\frac{1}{2}$ ,

(±$\sqrt{\frac{1-cosx}{2}}$)² = ($\frac{1}{2}$)²,

$\frac{1-cosx}{2}=\frac{1}{4} ,\\4-4cosx=2,\\-4cosx=-2,\\cosx=\frac{1}{2} .$

$cosx\geq \frac{1}{2}$

$x_{1} =\frac{\pi }{3}+2\pi *n\\\\x_{2} =\frac{5\pi }{3}+2\pi *n$

$\frac{1}{3}(6\pi *n-\pi ) \leq x\leq \frac{1}{3}(6\pi *n+\pi )$ , n∈Z

$ctg5x\geq 1$

$5x=\frac{\pi }{4} +\pi *n,\\x=\frac{\pi }{20} +\frac{ \pi *n}{5}.$

$\frac{\pi *n}{5} \leq x\leq \frac{\pi *n}{5}+\frac{\pi }{20}$ , n∈Z