Question

Решите тригометрическое уравнение:
cos3x-cos5x+cos7x-cos9x=0

Подробно​

Answer 1

Ответ:

$\boxed{\left [ \begin{array}{ccc} x = \pi n, n \in \mathbb Z \\ \\ x = \dfrac{\pi n}{2} , n \in \mathbb Z \\ \\ x = \dfrac{\pi n}{6}, n \in \mathbb Z \end{array}\right}$

Формулы:

$\cos \alpha - \cos \beta = -2 \sin \bigg (\dfrac{\alpha + \beta }{2} \bigg) \sin \bigg (\dfrac{\alpha - \beta }{2} \bigg)$

$\sin \alpha + \sin \beta = 2 \sin \bigg (\dfrac{\alpha + \beta }{2} \bigg) \cos \bigg (\dfrac{\alpha - \beta }{2} \bigg)$

$\sin x = 0 \Longrightarrow x = \pi n, n \in \mathbb Z$

Объяснение:

$\cos 3x - \cos 5x + \cos 7x - \cos 9x = 0|\cdot(-1)$

$\cos 5x - \cos 3x + \cos 9x - \cos 7x = 0$

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а)

$\cos 5x - \cos 3x = -2 \sin \bigg (\dfrac{5x + 3x }{2} \bigg) \sin \bigg (\dfrac{5x - 3x }{2} \bigg) = -2 \sin \bigg (\dfrac{8x }{2} \bigg) \sin \bigg (\dfrac{2x }{2} \bigg) =$

$= -2 \sin 4x \sin x$

б)

$\cos 9x - \cos 7x = -2 \sin \bigg (\dfrac{9x + 7x }{2} \bigg) \sin \bigg (\dfrac{9x - 7x}{2} \bigg) =-2 \sin \bigg (\dfrac{16x }{2} \bigg) \sin \bigg (\dfrac{2x }{2} \bigg) =$

$= -2 \sin 8x \sin x$

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$-2 \sin 4x \sin x -2 \sin 8x \sin x = 0| \cdot (-0,5)$

$\sin 4x \sin x + \sin 8x \sin x = 0$

$\sin x(\sin 8x + \sin 4x) = 0$

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в)

$\sin 8x + \sin 4x = 2 \sin \bigg (\dfrac{8x + 4x}{2} \bigg) \cos \bigg (\dfrac{8x - 4x}{2} \bigg) = 2 \sin \bigg (\dfrac{12x}{2} \bigg) \cos \bigg (\dfrac{4x}{2} \bigg) =$

$= 2 \sin 6x \sin 2x$

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$2 \sin 6x \sin 2x \sin x = 0|:2$

$\sin x \sin 2x \sin 6x = 0 \Longleftrightarrow \left [ \begin{array}{ccc} \sin x = 0 \\ \sin 2x = 0 \\ \sin 6x = 0 \end{array}\right$

$\left [ \begin{array}{ccc} x = \pi n, n \in \mathbb Z \\ 2x = \pi n, n \in \mathbb Z|:2 \\ 6x = \pi n, n \in \mathbb Z|:6 \end{array}\right$

$\left [ \begin{array}{ccc} x = \pi n, n \in \mathbb Z \\ \\ x = \dfrac{\pi n}{2} , n \in \mathbb Z \\ \\ x = \dfrac{\pi n}{6}, n \in \mathbb Z \end{array}\right$