Question

Найдите сумму корней уравнения 1+cos2x-2sin^2x=1 на промежутке [0,2pi]

Answer 1

$2 cos^{2}x-2(1- cos^{2}x)=1, cos^{2} x= \frac{3}{4};$
$cosx= \frac{ \sqrt{3} }{2}, x= +-\frac{ \pi }{6}+2 \pi n ,$n∈Z;
На $[0;2 \pi ] x_{1}= \frac{ \pi }{6}, x_{2}= \frac{11 \pi }{6};$
$cosx=- \frac{ \sqrt{3} }{2}, x=+- \frac{5 \pi }{6}+2 \pi k,$ k∈Z;
На $[0; 2 \pi ], x_{3} = \frac{5 \pi }{6}, x_{4}= \frac{7 \pi }{6};$
$x_{1} + x_{2}+ x_{3}+ x_{4}= \frac{ \pi }{6}+ \frac{ 11\pi }{6}+ \frac{5 \pi }{6}+ \frac{7 \pi }{6}=4 \pi .$