cos2x+sin^2x=0,25 + отобрать корни на отрезке [3pi;9pi/2]. Created by IOriOnI. algebra-ru

Отбор корней на отрезке: нет таких Ответ:

2]...

F1n4ek $\cos2x+\sin^2x=0.25;$

$\cos^2x-\sin^2x+\sin^2x=\frac{1}{4};$

$\cos^2x=\frac{1}{4};$

$a)\ \cosx=\frac{1}{2}; b)\ \cosx=-\frac{1}{2};$

$a)\ \cosx=\frac{1}{2};$

$x=\pm\arccos\frac{1}{2}+2\pi n,\ n\in Z;$

$x=\pm\frac{\pi}{3}+2\pi n,\ n\in Z;$

$b)\ \cosx=-\frac{1}{2};$

$x=\pm\arccos(-\frac{1}{2})+2\pi n,\ n\in Z;$

$x=\pm (\pi-\frac{\pi}{3})+2\pi n,\ n\in Z;$

$x=\pm\frac{2\pi}{3}+2\pi n,\ n\in Z;$

Отбор корней на отрезке: $[3\pi; \frac{9\pi}{2} ];$

$a)\ x = \pm\frac{\pi}{3}+2\pi n;$
$1)\ x=\frac{\pi}{3}+2\pi n, n(Z;$
$3\pi\leq\frac{\pi}{3}+2\pi n\leq\frac{9\pi}{2};$

$3\leq\frac{1}{3}+2n\leq\frac{9}{2};$

$3-\frac{1}{3}\leq 2n \leq \frac{9}{2}-\frac{1}{3};$

$\frac{8}{3}\leq 2n \leq \frac{25}{6};$

$16\leq12n\leq25;$
$n=2;$
$x=\frac{\pi}{3}+2\pi\bullet2= \frac{13\pi}{3};$

$2)\ x=-\frac{\pi}{3}+2\pi n,\ n\in Z;$

$3\pi \leq-\frac{\pi}{3}+2\pi n\leq\frac{9\pi}{2};$

$3\leq-\frac{1}{3}+2n\leq\frac{9}{2};$

$20\leq12n\leq29;$
$n=2;$
$x=-\frac{\pi}{3}+2\pi\bullet2 = -\frac{\pi}{3}+4\pi = \frac{11\pi}{3};$

$b)\ x=\pm \frac{2\pi}{3}=2\pi n,\ n\in Z;$

$1)\ x=\frac{2\pi}{3}+2\pi n,\ n\in Z;$

$3\pi\leq\frac{2\pi}{3}+2\pi n\leq\frac{9\pi}{2};$

$3\leq\frac{2}{3}+2n\leq\frac{9}{2};$

$18\leq4+12n\leq\27;$

$14\leq12n\leq23;$
нет таких $n\in Z.$

$2)\ x=-\frac{2\pi}{3}+2\pi n,\ n\in Z;$

$3\pi\leq-\frac{2\pi}{3}+2\pi n\leq\frac{9\pi}{2};$

$3\leq-\frac{2}{3}+2 n\leq\frac{9}{2};$

$22\leq12n\leq31;$
$n= 2;$

$x=-\frac{2\pi}{3}+2\pi*2=-\frac{2\pi}{3}+4\pi=\frac{10\pi}{3};$

Ответ: $1)x=б\frac{\pi}{3}+2\pi n,$ $x=б\frac{2\pi}{3}+2\pi n,\ n\in Z;$
$2)\ x=\frac{10\pi}{3},$
$x=\frac{11\pi}{3},$
$x=\frac{13\pi}{3}.$

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