Question

найдите корни уравнения (тригонометрия)
$\cos \frac{\pi(2x - 1)}{3} = \frac{1}{2}$
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Answer 1

π(2х-1)/3=±π/3+2πn; n∈Z

2x-1=±1+6n n∈Z

2x=±1+1+6n n∈Z

х=±0.5+0.5+3n n∈Z

х₁=1+3n n∈Z

х₂=3n n∈Z

Answer 2

$Cos\frac{\pi(2x-1) }{3}=\frac{1}{2}\\\\\frac{\pi(2x-1)}{3}=\pm\frac{\pi }{3}+2\pi n,n\in Z\\\\\pi(2x-1)=\pm\pi+6\pi n,n\in Z\\\\2x-1=\pm1+6n,n\in Z\\\\1)2x=2+6n,n\in Z\\\\x_{1}=1+3n,n\in Z\\\\2)2x=6n,n\in Z\\\\x_{2}=3n,n\in Z$