Предмет: Алгебра, автор: Аноним

Решите уравнение:
1) IsinxI=IcosxI
2) $sqrt{3} ctgx=$ 2IcosxI

Ответы

Автор ответа: Artem112

0

1.
$|sin x|=|cos x|$
Два модуля равны в случае, если подмодульные выражения равны или они противоположны. Получаем совокупность:
$left[begin{array}{l} sin x=cos x \ sin x=-cos x end{array}$
$left[begin{array}{l} mathrm{tg} x=1\ mathrm{tg} x=-1 end{array}$
$left[begin{array}{l} x= frac{ pi }{4} + pi k \ x=- frac{ pi }{4} + pi k end{array}$
$x=pm frac{ pi }{4} + pi k , kin Z$

2.
$sqrt{3}cdot mathrm{ctg}x= 2|cos x| \ sqrt{3}cdot dfrac{cos x}{sin x} = 2|cos x|$
Если $cos x geq 0$ :
$sqrt{3}cdot frac{cos x}{sin x} = -2cos x \ cos x =0 \ x_1= frac{ pi }{2}+ pi n, nin Z \ sqrt{3}cdot frac{1}{sin x} = 2 \ sin x = frac{ sqrt{3} }{2} \ x_2= frac{ pi }{3} +2 pi n, nin Z; x_3= frac{ 2pi }{3} +2 pi n, nin Z$
Учитывая ограничение, при котором раскрыт модуль, третья серия корней отбрасывается.
Если $cos x textless 0$ :
$sqrt{3}cdot frac{cos x}{sin x} = -2cos x \ sqrt{3}cdot frac{1}{sin x} = -2 \ sin x =- frac{ sqrt{3} }{2} \ x_4= frac{ 4pi }{3} +2 pi n, nin Z; x_5= frac{ 5pi }{3} +2 pi n, nin Z$
Учитывая ограничение, при котором раскрыт модуль, пятая серия корней отбрасывается.
Оставшиеся корни:
$left[begin{array}{l} x_1= frac{ pi }{2} + pi n \ x_2= frac{ pi }{3} +2 pi n \ x_4= frac{ 4pi }{3} +2 pi n end{array}$
$left[begin{array}{l} x_1= frac{ pi }{2} + pi n \ x_2= frac{ pi }{3} + pi n end{array}, nin Z$

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