3.4. 1) $\sin\left(x-\frac{\pi}{3}\right)=0{,}5$; 2) $\cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt3}{2}$; 3) $ g\left(2x-\frac{\pi}{6}\right)=1$; 4) $\ctg\left(\frac{x}{2}+\frac{\pi}{8}\right)=\sqrt3$.

Question

Вопрос: $Фото задания: 3.4. 1) $\sin\left(x-\frac{\pi}{3}\right)=0{,}5$; 2) $\cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt3}{2}$; 3) \(…$ Фото задания

3.4. 1) $\sin\left(x-\frac{\pi}{3}\right)=0{,}5$; 2) $\cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt3}{2}$; 3) $ g\left(2x-\frac{\pi}{6}\right)=1$; 4) $\ctg\left(\frac{x}{2}+\frac{\pi}{8}\right)=\sqrt3$.

Accepted Answer

Барлық шешімдерде $k\in\mathbb Z$.

$u=x-\frac{\pi}{3}$ десек, $\sin u=\frac12$. Демек, $u=\frac{\pi}{6}+2\pi k$ немесе $u=\frac{5\pi}{6}+2\pi k$, сондықтан $x=\frac{\pi}{2}+2\pi k$ немесе $x=\frac{7\pi}{6}+2\pi k$.
$u=x+\frac{\pi}{4}$ десек, $\cos u=-\frac{\sqrt3}{2}$. Демек, $u=\frac{5\pi}{6}+2\pi k$ немесе $u=\frac{7\pi}{6}+2\pi k$, сондықтан $x=\frac{7\pi}{12}+2\pi k$ немесе $x=\frac{11\pi}{12}+2\pi k$.
$2x-\frac{\pi}{6}=\frac{\pi}{4}+\pi k$. Осыдан $x=\frac{5\pi}{24}+\frac{\pi k}{2}$.
$\frac{x}{2}+\frac{\pi}{8}=\frac{\pi}{6}+\pi k$. Осыдан $x=\frac{\pi}{12}+2\pi k$.

3.4. 1) \(\sin\left(x-\frac{\pi}{3}\right)=0{,}5\); 2) \(\cos\left(x+\frac{\pi}{4}\right)=-\frac{\sqrt3}{2}\); 3) \( g\left(2x-\frac{\pi}{6}\right)=1\); 4) \(\ctg\left(\frac{x}{2}+\frac{\pi}{8}\right)=\sqrt3\).

Ответ: