Question

Решите уравнение :
cos 2x + 10 sin x - 9 = 0. Найдите решение на отрезке [-π;π].

Answer 1

cos(2x)=1-2sin^2(x) => 1-2sin^2(x)+10sin(x)-9=0 => 2sin^2(x)-10sin(x)+8=0 => sin^2(x)-5sin(x)+4=0. Пусть y=sin(x), |y|<=1 => y^2-5y+4=0 => по теореме Виета y1+y2=5, y1y2=4 => y1=1, y2=4 => y=1 (|y|<=1) => sin(x)=1 => x=pi/2+2pi*n, nEZ. -pi<=pi/2+2pi*n<=pi => -1<=1/2+2n<=1 => -3/2<=2n<=1/2 => -3/4<=n<=1/4 => n=0, откуда x=pi/2+2pi*0=pi/2. Ответ: x=pi/2+2pi*n, nEZ; pi/2.