3.6. 1) $\sqrt2\sin\left(x-\frac{\pi}{6}\right)=1$; 2) $2\cos\left(2x+\frac{\pi}{3}\right)=\sqrt3$; 3) $\sqrt3\cdot g\left(\frac{x}{2}-\frac{\pi}{4}\right)=1$; 4) $\ctg\left(2x+\frac{\pi}{3}\right)=\sqrt3$.

Question

Вопрос: $Фото задания: 3.6. 1) $\sqrt2\sin\left(x-\frac{\pi}{6}\right)=1$; 2) $2\cos\left(2x+\frac{\pi}{3}\right)=\sqrt3$; 3) \(\sqrt3\cdot…$ Фото задания

3.6. 1) $\sqrt2\sin\left(x-\frac{\pi}{6}\right)=1$; 2) $2\cos\left(2x+\frac{\pi}{3}\right)=\sqrt3$; 3) $\sqrt3\cdot g\left(\frac{x}{2}-\frac{\pi}{4}\right)=1$; 4) $\ctg\left(2x+\frac{\pi}{3}\right)=\sqrt3$.

Accepted Answer

Барлық шешімдерде $k\in\mathbb Z$.

$\sin\left(x-\frac{\pi}{6}\right)=\frac{\sqrt2}{2}$. Сондықтан $x-\frac{\pi}{6}=\frac{\pi}{4}+2\pi k$ немесе $\frac{3\pi}{4}+2\pi k$. Демек, $x=\frac{5\pi}{12}+2\pi k$ немесе $x=\frac{11\pi}{12}+2\pi k$.
$\cos\left(2x+\frac{\pi}{3}\right)=\frac{\sqrt3}{2}$. Сондықтан $2x+\frac{\pi}{3}=\pm\frac{\pi}{6}+2\pi k$, демек $x=-\frac{\pi}{12}+\pi k$ немесе $x=-\frac{\pi}{4}+\pi k$.
$ g\left(\frac{x}{2}-\frac{\pi}{4}\right)=\frac{1}{\sqrt3}$. Сондықтан $\frac{x}{2}-\frac{\pi}{4}=\frac{\pi}{6}+\pi k$, демек $x=\frac{5\pi}{6}+2\pi k$.
$2x+\frac{\pi}{3}=\frac{\pi}{6}+\pi k$. Демек, $x=-\frac{\pi}{12}+\frac{\pi k}{2}$.

3.6. 1) \(\sqrt2\sin\left(x-\frac{\pi}{6}\right)=1\); 2) \(2\cos\left(2x+\frac{\pi}{3}\right)=\sqrt3\); 3) \(\sqrt3\cdot g\left(\frac{x}{2}-\frac{\pi}{4}\right)=1\); 4) \(\ctg\left(2x+\frac{\pi}{3}\right)=\sqrt3\).

Ответ: