Question

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

Answer 1

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

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

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

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

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

Ответ:

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