СРОЧНО ПОМОГИТЕ РЕШИТЬ!!! cos3x + sin2x

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Nikkaria @Nikkaria

July 2022 1 4 Report

СРОЧНО ПОМОГИТЕ РЕШИТЬ!!!
cos3x + sin2x - cos x =0

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nikebod313

$\cos 3x + \sin 2x - \cos x = 0$

$\cos 3x - \cos x + \sin 2x = 0$

По формуле $\cos \alpha - \cos \beta = -2\sin \dfrac{\alpha - \beta } {2} \sin \dfrac{\alpha + \beta } {2}$ имеем:

$-2\sin x \sin 2x + \sin 2x = 0$

$\sin2x (-2\sin x + 1) = 0$

$\left[\begin{array}{ccc}\sin 2x = 0 \ \ \ \ \ \ \ \\-2\sin x + 1 = 0\\\end{array}\right$

$\left[\begin{array}{ccc} 2x = \pi k, \ k \in Z \\\sin x = \dfrac{1}{2} \ \ \ \ \ \ \ \ \\\end{array}\right$

$\left[\begin{array}{ccc} x = \dfrac{\pi k}{2} , \ k \in Z \ \ \ \ \ \ \ \ \ \ \ \ \ \\\ x = (-1)^{n}\dfrac{\pi}{6} + \pi n, \ n \in Z \\\end{array}\right$

Ответ: $\dfrac{\pi k}{2}; \ (-1)^{n}\dfrac{\pi}{6} + \pi n; \ k, \ n \in Z$

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