Найдите корни уравнения, принадлежащие отрезку [π; 2π]. [tex] cos^2x=\sqrt{2} sin(2x)-sin^2x [/tex]. Created by 0vvl. matematika-ru

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Удачник66

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√2*sin(2x) = cos^2 x + sin^2 x = 1

sin(2x) = 1/√2

2x = (-1^n)*pi/4 + pi*n

x = (-1)^n*pi/8 + pi/2*n

Отрезку [pi; 2pi] принадлежат корни:

x1(n=2) = pi/8 + pi = 9pi/8; x2(n=3) = -pi/8 + 3pi/2 = 11pi/8

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InvisorTech "Найдите КОРНИ уравнения, принадлежащие ОТРЕЗКУ [π; 2π]".

Удачник66 Извиняюсь, не заметил этого условия. Это корни x1=pi+pi/8=9pi/8, x2=3pi/2-pi/8=11pi/8

InvisorTech

$\cos^{2}x = \sqrt{2}\sin 2x - \sin^{2}x \\ \\ 1 = \sqrt{2} \sin 2x \\ \\ \sin 2x = \dfrac{\sqrt{2}}{2} \\ \\ $\left[ \begin{gathered} 2x = \dfrac{\pi}{4}+ 2\pi n, n \in Z \\ 2x = \dfrac{3\pi}{4} + 2\pi k, k \in Z \\ \end{gathered} \right.$$

$\left[ \begin{gathered} x = \dfrac{\pi}{8}+ \pi n, n \in Z \\ x = \dfrac{3\pi}{8} + \pi k, k \in Z \\ \end{gathered} \right.$

Корни уравнения (1):

$\pi + \dfrac{\pi}{8} = \dfrac{9\pi}{8} \\ \\ \dfrac{3\pi}{2} - \dfrac{\pi}{8} = \dfrac{11\pi}{8}$

Ответ: 9π/8, 11π/8

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