Question

решить уравнения:

1)sin3x=cos3x найти корни уравнения на отрезке(0,4)

2)sin²x-2cosx+2=0

Answer 1

$1) sin3x = cos3x |:cos3x \ \ tg3x = 1 \ \ 3x = dfrac{ pi }{4} + pi n, n in Z \ \ x = dfrac{pi}{12} + dfrac{ pi n }{3} , n in Z \ \ 0 textless dfrac{pi}{12} + dfrac{ pi n }{3} textless 4, n in Z \ \ -dfrac{pi}{12} textless dfrac{ pi n }{3} textless 4 - dfrac{ pi}{12}, n in Z \ \ - pi textless 4 pi n textless 48 - pi , n in Z \ \ n = 0; 1; 2; 3$

$x_1 = dfrac{pi}{12} \ \ x_2 = dfrac{pi}{12} + dfrac{pi}{3} = dfrac{5 pi}{12} \ \ x_2 = dfrac{pi}{12} +dfrac{2pi}{3} = dfrac{3 pi}{4} \ \ x_4 = dfrac{pi}{12} + pi = dfrac{13pi}{12} \ \ boxed{OTBET: x = dfrac{pi}{12} ; dfrac{5 pi}{12} ; dfrac{3 pi}{4} ; dfrac{13pi}{12}. }$


$2. sin^2x - 2cosx + 2 = 0 \ \ 1 - cos^2x - 2cosx + 2 = 0 \ \ -cos^2x - 2cosx + 3 = 0 \ \ cos^2x - 2cosx - 3 = 0 \ \ cos^2x - 2cosx + 1 - 4 = 0 \ \ (cosx - 1)^2 - 4 = 0 \ \ (cosx - 1 - 2)(cosx - 1 + 2) = 0 \ \ (cosx - 2)(cosx + 1) = 0 \ \ cosx = 2 - ne ud. \ \ cosx = -1 \ \ boxed{ x = pi + 2 pi n, n in Z}$