Question

Найдите корни заданного уравнения на заданном промежутке: cos x=√3/2,x∈[0; 2p]

Answer 1

$\cos x=\frac{\sqrt3}2\\x_1=-\frac\pi6+2\pi n,\;\;x_2=\frac\pi6+2\pi n\;n\in\mathbb{Z}\\a)\;0\leq-\frac\pi6+2\pi n\leq2\pi\\\frac\pi6\leq2\pi n\leq\frac{13\pi}6\\\frac1{12}\leq n\leq\frac{13}6\\n\in\mathbb{Z}\Rightarrow n=1\Rightarrow x_1=-\frac\pi 6+2\pi=\frac{11\pi}6\\\\b)\;0\leq\frac\pi6+2\pi n\leq2\pi\\-\frac\pi6\leq2\pi n\leq\frac{11\pi}6\\-\frac1{12}\leq n\leq\frac{11}6\\n\in\mathbb{Z}\Rightarrow n=0\Rightarrow x_2=\frac\pi6$