Question

Решите уравнение:
sin(x+$\frac{5\pi }{2}$)+2sin(2x+$\frac{\pi }{2}$)=cos(3x+π)

Answer 1

Ответ:

$\sin(x + \frac{5\pi}{2} ) + 2 \sin(2x + \frac{\pi}{2} ) = \cos(3x + \pi)$

$\cos(x) + 2 \cos(2x) = - \cos(3x)$

$\cos(x) + 2 \cos(2x) + \cos(3x) = 0$

$2 \cos(2x) \cos( - x) + 2 \cos(2x) = 0$

$2 \cos(2x) ( \cos( - x) + 1) = 0$

$2 \cos(2x) ( \cos(x) + 1) = 0$

$\cos(2x) ( \cos(x) + 1) = 0$

$\cos(2x) = 0 \\ \cos(x) + 1 = 0$

$x = \frac{\pi}{4} + \frac{k\pi}{2} \\ x = \pi + 2k\pi$

Answer 2

Ответ:

$sin(x+\dfrac{5\pi }{2})+2sin(2x+\dfrac{\pi}{2})=cos(3x+\pi )\\\\cosx+2cos2x=-cos3x\\\\(cosx+cos3x)+2cos2x=0\\\\2\cdot cos\dfrac{x+3x}{2}\cdot cos\dfrac{3x-x}{2}+2\, cos2x=0\\\\2\cdot cos2x\cdot cosx+2\, cos2x=0\\\\2\, cos2x\cdot (cosx+1)=0\\\\a)\ \ cos2x=0\ \ ,\ \ 2x=\dfrac{\pi}{2}+\pi n\ ,\ \ x=\dfrac{\pi}{4}+\dfrac{\pi n}{2}\ ,\ n\in Z\\\\b)\ \ cosx+1=0\ \ ,\ \ cosx=-1\ \ ,\ \ \ x=\pi +2\pi k\ ,\ k\in Z\\\\Otvet:\ \ x_1=\dfrac{\pi}{4}+\dfrac{\pi n}{2}\ ,\ \ x_2=\pi (2k+1)\ \ ,\ n,k\in Z\ .$