Question

Помогите решить тригонометрию
Корень из 0,5 cosx = sin x/2

Answer 1

$sqrt{frac{1}{2}cos x} = sin left(frac{x}{2}right)\left{ begin{aligned}&frac{1}{2}cos x geqslant 0 \&sin left(frac{x}{2}right) geqslant 0 \ &frac{1}{2} cos x = { sin}^2 left(frac{x}{2}right) end{aligned} right. \ left{ begin{aligned}&cos x geqslant 0 \&sin left(frac{x}{2}right) geqslant 0 \ &cos x - 2 :{ sin}^2 left(frac{x}{2}right) = 0end{aligned} right.$

По формуле $cos x = 1 - 2 : sin^2 left(frac{x}{2}right)$

$left{ begin{aligned}&cos x geqslant 0 \&sin left(frac{x}{2}right) geqslant 0 \ &1 - 4 :{ sin}^2 left(frac{x}{2}right) = 0end{aligned} right.$

Рассмотрим $1 - 4 :{ sin}^2 left(frac{x}{2}right) = 0$

$1 - 4 :{ sin}^2 left(frac{x}{2}right) = 0\{ sin}^2 left(frac{x}{2}right) = frac{1}{4}\sin left(frac{x}{2}right) = pm frac{1}{2}$

Вернёмся к системе

$left{ begin{aligned}&cos x geqslant 0 \&sin left(frac{x}{2}right) geqslant 0 \ &sin left(frac{x}{2}right) = pm frac{1}{2}end{aligned} right.\left{ begin{aligned}&cos x geqslant 0\ &sin left(frac{x}{2}right) = frac{1}{2}end{aligned} right.\left{ begin{aligned}&cos x geqslant 0\ &left[ begin{aligned} &frac{x}{2}=frac{pi}{6} +2pi k_1, : k_1 in mathbb{Z}\&frac{x}{2}=frac{5pi}{6} +2pi k_2, : k_2 in mathbb{Z}end{aligned} right. end{aligned} right.\left{ begin{aligned}&cos x geqslant 0\ &left[ begin{aligned} &x=frac{pi}{3} +4pi k_1, : k_1 in mathbb{Z}\&x=frac{5pi}{3} +4pi k_2, : k_2 in mathbb{Z}end{aligned} right. end{aligned} right. \ left[ begin{aligned} &x=frac{pi}{3} +4pi k_1, : k_1 in mathbb{Z}\&x=frac{5pi}{3} +4pi k_2, : k_2 in mathbb{Z}end{aligned} right.$

Ответ:

$left[ begin{aligned} &x=frac{pi}{3} +4pi k_1, : k_1 in mathbb{Z}\&x=frac{5pi}{3} +4pi k_2, : k_2 in mathbb{Z}end{aligned} right. \Downarrow\ x=pmfrac{pi}{3} +4pi k, : k in mathbb{Z}$