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

========

Answer 2

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

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