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July 2022 1 47 Report

Определить количество корней уравнения на отрезке [tex][ \frac{- \pi }{8} ; \frac{ \pi }{8} ][/tex]

Само уравнение:
[tex]2cos (2x+\frac{ \pi}{3} )*sin (3x- \frac{ \pi}{4}) - sin (x - \frac{ 7 \pi}{12}) - cos ( \frac{ 5 \pi}{12} - 11x) = 0 [/tex]

Answers & Comments

Матов

Verified answer

$2cos(2x+\frac{\pi}{3})*sin(3x-\frac{\pi}{4})-sin(x-\frac{7\pi}{12})-cos(\frac{5\pi}{12}-11x)=0\\\\
sin(x-\frac{7\pi}{12})=-cos(\frac{\pi}{12}-x)\\\\
2cos(2x+\frac{\pi}{3})*sin(3x-\frac{\pi}{4})+cos(\frac{\pi}{12}-x)-cos(\frac{5\pi}{12}-11x)=0\\\\
$
по формуле произведение
$2cos(2x+\frac{\pi}{3})*sin(3x-\frac{\pi}{4})=sin(5x+\frac{\pi}{12})-cos(\frac{\pi}{12}-x)\\\\
sin(5x+\frac{\pi}{12})-cos(\frac{\pi}{12}-x)+cos(\frac{\pi}{12}-x)-cos(\frac{5\pi}{12}-11x)=0\\\\
sin(5x+\frac{\pi}{12})-cos(\frac{5\pi}{12}-11x)=0\\\\
cos(\frac{5\pi}{12}-11x)=sin(11x+\frac{\pi}{12})\\\\
sin(5x+\frac{\pi}{12})-sin(11x+\frac{\pi}{12})=0\\\\
-2sin3x*cos(8x+\frac{\pi}{12})=0\\\\
x=\frac{\pi\*n}{3}\\\\ 
x=\frac{\pi\*n}{8}-\frac{7\pi}{96}$
Ответ 3

1 votes Thanks 1

Answers & Comments

Verified answer

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