Question

Найти все корни уравнения cosx + sinx = 1/$cosx + sinx = \frac{1}{\sqrt[2]{2}}$ на интервале x от 0 до pi

Answer 1

$\cos x + \sin x = \dfrac{1}{\sqrt{2}} \ \ \ \Big| : \sqrt{2}$

$\dfrac{1}{\sqrt{2}} \cos x + \dfrac{1}{\sqrt{2}}\sin x = \dfrac{1}{2}$

$\sin \dfrac{\pi}{4} \cos x + \cos \dfrac{\pi}{4} \sin x = \dfrac{1}{2}$

$\sin \left(\dfrac{\pi}{4} + x \right) = \dfrac{1}{2}$

$\dfrac{\pi}{4} + x = (-1)^{n} \arcsin \dfrac{1}{2} + \pi n, \ n \in Z$

$\dfrac{\pi}{4} + x = (-1)^{n} \dfrac{\pi}{6} + \pi n, \ n \in Z$

$x = -\dfrac{\pi}{4} + (-1)^{n} \dfrac{\pi}{6} + \pi n, \ n \in Z$

Если $n = 0$, то $x = -\dfrac{\pi}{4} + (-1)^{0} \dfrac{\pi}{6} + \pi \cdot 0 = -\dfrac{\pi}{12} \notin (0; \ \pi)$

Если $n = 1$, то $x = -\dfrac{\pi}{4} + (-1)^{1} \dfrac{\pi}{6} + \pi \cdot 1 = \dfrac{7\pi}{12} \in (0; \ \pi)$

Если $n =2$, то $x = -\dfrac{\pi}{4} + (-1)^{2} \dfrac{\pi}{6} + \pi \cdot 2 = \dfrac{23\pi}{12} \notin (0; \ \pi)$

Если $n = -1$, то $x = -\dfrac{\pi}{4} + (-1)^{-1} \dfrac{\pi}{6} + \pi \cdot (-1) = -\dfrac{17\pi}{12} \notin (0; \ \pi)$

Ответ: $x = \dfrac{7\pi}{12}$