Алгебра: Cos2x+sin^2x=0,25 + отобрать корни на отрезке [3pi; 9pi/2]

Отбор корней на отрезке: нет таких ответ: ...

Cos2x+sin^2x=0,25 + отобрать корни на отрезке [3pi; 9pi/2]

Ответы

148625 24.05.2020 14:56

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

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

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

$<spana)\ \cosx=\frac{1}{2}; b)\ \cosx=-\frac{1}{2}</span;$

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

$<spanx=\pm\arccos\frac{1}{2}+2\pi n,\ n\in Z;</span$

$<spanx=\pm\frac{\pi}{3}+2\pi n,\ n\in Z;</span$

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

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

$<spanx=\pm (\pi-\frac{\pi}{3})+2\pi n,\ n\in Z</span;$

$<spanx=\pm\frac{2\pi}{3}+2\pi n,\ n\in Z;</span$

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

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

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

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

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

$16\leq12n\leq25;$
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 $<span<span<spann\in Z.</span$

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

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

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

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

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

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