Question

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

Answer 1

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

Answer 2

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

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

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

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

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.