Question

А) решите уравнение: 2-sin^2x=cos^2x+cos(п/2-3x) б) укажите корни, принадлежащие промежутку [-п/2; п/2]

Answer 1

$2 - sin^2x = cos^2x + cos \bigg ( \dfrac{ \pi }{2} - 3x \bigg ) \\ \\ 2 - cos^2x - sin^2x = cos \bigg ( \dfrac{ \pi }{2} - 3x \bigg ) \\ \\ 2 - 1 = sin3x \\ \\ sin3x = 1 \\ \\ 3x = \dfrac{ \pi }{2} + 2 \pi n, \ n \in Z \\ \\ x = \dfrac{\pi}{6} + \dfrac{2 \pi n}{3} , \ n \in Z \\ \\ - \dfrac{ \pi }{2} \leq \dfrac{\pi}{6} + \dfrac{2 \pi n}{3} \leq \dfrac{ \pi }{2} , \ n \in Z \\ \\ -3 \pi \leq \pi + 4 \pi n \leq 3 \pi , \ n \in Z \\ -3 \leq 1 + 4n \leq 3, n \in Z \\$
$-4 \leq 4n \leq 2, \ n \in Z \\ \\ -1 \leq n \leq \dfrac{1}{2}, \ n \in Z \\ \\ n = -1; \ 0. \\ x_1 = \dfrac{ \pi }{6} - \dfrac{2 \pi }{3} = \dfrac{ \pi }{6} - \dfrac{4 \pi }{6} = - \dfrac{\pi }{2} \\ \\ x_2 = \dfrac{ \pi }{6} \\ \\ OTBET: \boxed{\ x = \dfrac{\pi }{2} ; \ \dfrac{ \pi }{6}.}$

Answer 2

2-sin²x-cos²x=sin3x
2-(sin²x+cos²x)=sin3x
sin3x=2-1
sin3x=1
3x=π/2+2πk
x=π/6+2πk/3,k∈z
-π/2≤π/6+2πk/3≤π/2
-3≤1+4k≤3
-4≤4k≤2
-1≤k≤0,5
k=-1⇒x=π/6-2π/3=-π/2
k=0⇔x=π/6