Question

решить уравнения:1)sin3x=cos3x найти корни уравнения на отрезке(0,4)2)sin²x-2cosx+2=0

Answer 1

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

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


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