Question

Реши уравнение cos(2π+8x)=sqrt{2}/2и выбери корни из отрезка [π/2;π].

Answer 1

$cos(2pi + 8x) = frac{ sqrt{2} }{2} \ cos(8x) = frac{ sqrt{2} }{2} \ left[ begin{gathered} 8x = frac{pi}{4} + 2pi n \8x = - frac{pi}{4} + 2pi n end{gathered} right. \ left[ begin{gathered} x = frac{pi}{32} + frac{pi}{4}n,: n in mathbb Z \ x = - frac{pi}{32} + frac{pi}{4} m, : m in mathbb Z end{gathered} right.$

Отбираем корни:

$frac{pi}{2} leqslant frac{pi}{32} + frac{pi}{4} n leqslant pi \ frac{1}{2} - frac{1}{32} leqslant frac{1}{4} n leqslant 1 - frac{1}{32} \ frac{15 times 4}{32} leqslant n leqslant frac{31 times 4}{32} \ frac{15}{8} leqslant n leqslant frac{31}{8} \ n = {1,2,3 } \ left[ begin{gathered}x_{1} = frac{pi}{32} + frac{pi}{4} = frac{9pi}{32} \ x_{2} = frac{pi}{32} + frac{2pi}{4} = frac{17pi}{32} \ x_{3} = frac{pi}{32} + frac{3pi}{4} = frac{25pi}{32} end{gathered} right.$

$frac{pi}{2} leqslant - frac{pi}{32} + frac{pi}{4} m leqslant pi \ frac{1}{2} + frac{1}{32} leqslant frac{1}{4} m leqslant 1 + frac{1}{32} \ frac{17 times 4}{32} leqslant m leqslant frac{33 times 4}{32} \ frac{17}{8} leqslant m leqslant frac{33}{8} \ m = {2,3,4 } \ left[ begin{gathered}x_{1} = - frac{pi}{32} + frac{pi}{2} = frac{15pi}{32} \ x_{2} = - frac{pi}{32} + frac{3pi}{4} = frac{23pi}{32} \ x_{3} = - frac{pi}{32} + frac{4pi}{4} = frac{31pi}{32} end{gathered} right.$