Question

1)3cos10x–8cos5x+5=0 2)Промежуток [π/6,π/2] Объясните 2часть поподробней

Answer 1

1)
$3cos10x-8cos5x+5=0\\6cos^25x-8cos5x+2=0\\2(cos5x-1)(3cos5x-1)=0\\\\x={2\pi k\over5},k\in Z\\x={\pm arccos{1\over3}+2\pi k\over5},k\in Z$

2)
${0\pi\over5}\ \textless \ {\pi\over6}\ \textless \ {2\pi\over5}\ \textless \ {\pi\over2}\ \textless \ {4\pi\over5}\\\\arccos{1\over3}\ \textless \ {\pi\over6}\\\\{arccos{1\over3}\over5}\ \textless \ {\pi\over6}\ \textless \ {2\pi\over5}-{\pi\over30}\ \textless \ {2\pi\over5}-{arccos{1\over3}\over5}}\ \textless \ {arccos{1\over3}\over5}+{2\pi\over5}\ \textless \ \\\ \textless \ {\pi\over30}+{2\pi\over5}\ \textless \ {\pi\over2}\ \textless \ {4\pi\over5}-{\pi\over30}\ \textless \ {4\pi\over5}-{arccos{1\over3}\over5}\\\\x\in\{{2\pi\over5},{2\pi\over5}-{arccos{1\over3}\over5},{2\pi\over5}+{arccos{1\over3}\over5}\}$