Question

Решить уравнение: cos3x+sin2x=sin4x на [0; pi]cos3x+sin2x=sin4x на [0; pi]cos3x+sin2x=sin4x на [0; pi]

Answer 1

$cos3x + sin2x = sin4x \\ \\ cos3x + sin2x - sin4x = 0 \\ \\ cos3x + sin \dfrac{2x - 4x}{2} \cdot cos \dfrac{2x + 4x}{2} = 0 \\ \\ cos3x - sinx \cdot cos3x = 0 \\ \\ cos3x(1 - sinx) = 0 \\ \\ cos3x = 0 \\ \\ 3x = \dfrac{ \pi }{2} + \pi n, \ n \in Z \\ \\ \boxed{x = \dfrac{ \pi }{6} + \dfrac{ \pi n}{3} , \ n \in Z} \\ \\ sinx = 1 \\ \\ \boxed{x = \dfrac{ \pi }{2} + 2 \pi k, \ k \in Z}$

$0 \leq \dfrac{ \pi }{6} + \dfrac{ \pi n}{3} \leq \pi , \ n \in Z \\ \\ 0 \leq \pi + 2 \pi n \leq 6 \pi , \ n \in Z \\ \\ 0 \leq 1 + 2n \leq 6, \ n \in Z \\ \\ -1 \leq 2n \leq 4, \ n \in Z \\ \\ n = 0; \ 1; \ 2.$

$x = \dfrac{ \pi }{6} + \dfrac{ \pi n}{3} , \ n \in Z \\ \\ x_1 = \dfrac{ \pi }{6} \\ \\ x_2 = \dfrac{ \pi }{6} + \dfrac{ \pi }{3}= \dfrac{ \pi }{2} \\ \\ x_3 = \dfrac{ \pi }{6} + \dfrac{2 \pi }{3} = \dfrac{ 5\pi }{6}$

$0 \leq \dfrac{ \pi }{2} + 2 \pi k \leq \pi , \ k \in Z \\ \\ 0 \leq 1 + 4k \leq 2, \ k \in Z \\ \\ -1 \leq 4k \leq 1, \ k \in Z \\ \\ k = 0 \\ \\ x = \dfrac{ \pi }{2}$

Ответ: $\boxed{ x = \dfrac{ \pi }{6}; \ \dfrac{ \pi }{2}; \ \dfrac{ 5\pi }{6}. }$