Question

Количество различных корней уравнения на промежутке [0;$pi$]

Answer 1

$sin3x+sin7x=2sin5x\\2sin5xcdot cos2x=2sin5x\\2sin5x(cos2x-1)=0\\sin5x=0; ,; ; 5x=pi n,; ; x=frac{pi n}{5}; ,; nin Z\\cos2x=1; ,; ; 2x=2pi k,; ; x=pi k; ,; kin Z\\ xin [, 0,pi , ]; ; :; ; ; x=0; ,frac{pi}{5}; ,; frac{2pi }{5}; ,; frac{3pi }{5}; ,; frac{4pi }{5}; ,; pi ; .$

Answer 2

sin 3x + sin 7x = 2·sin 5x
2·sin 5x·cos 2x = 2·sin 5x
2·sin 5x·(cos 2x - 1) = 0

$left[begin{array}{ccc}sin 5x =0\cos2x=1end{array}$

$left[begin{array}{ccc}5x = pi n, n E Z\2x= pi k,kEZend{array}$

$left[begin{array}{ccc}x = frac{ pi n}{5} , n E Z\x=pi k,kEZend{array}$

x = 0, $frac{ pi}{5}, frac{2 pi}{5}, frac{3 pi}{5}, frac{4 pi}{5}$, π
Количество корней на промежутке [0; π] - 6.