Question

Найдите решение уравнения на промежутке cos2x/5=0, x∈[π, 3/2π]

Answer 1

cos(2x/5)=0

2x/5=π/2+kπ, k∈Z

2x=5π/2+5kπ, k∈Z

x=5π/4+5kπ/2, k∈Z

π≤x≤3π/2

π≤5π/4+5kπ/2≤3π/2

k=0 : π≤5π/4≤3π/2

x=5π/4

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Answer 2

$cos\frac{2x}{5}=0\\\\\frac{2x}{5}=\frac{\pi }{2}+\pi n\; ,\; n\in Z\\\\x=\frac{5\pi}{4}+ \frac{5\pi n}{2} \; ,\; n\in Z\\\\x\in [\, \pi ;\frac{3\pi}{2}\, ]:\; \; \pi \leq \frac{5\pi }{4}+\frac{5\pi n}{2}\leq \frac{3\pi }{2}\\\\\pi -\frac{5\pi }{4}\leq \frac{5\pi n}{2} \leq \frac{3\pi }{2}-\frac{5\pi }{4}\\\\-\frac{\pi }{4}\leq \frac{5\pi n}{2}\leq \frac{\pi }{4}\\\\-\frac{2\pi}{5\pi \cdot 4}\leq n\leq \frac{2\pi }{5\pi \cdot 4}\\\\-\frac{1}{10} \leq n\leq \frac{1}{10}\; ,\; n\in Z\; \; \Rightarrow \; \; n=0$

$n=0\; \; :\; \; x=\frac{5\pi }{4}\in [\, \pi ;\frac{3\pi}{2}\, ]$