Question

СРОЧНО‼️ДАЮ 40Б Розв'язання рiвняння: 1) sin 5x - sin x = 0; 2) cos2x + 5sinx - 4 = 0 ​

Answer 1

$\displaystyle\bf\\1)\\\\Sin5x-Sinx=0\\\\\\2Sin\frac{5x-x}{2} Cos\frac{5x+x}{2} =0\\\\\\2Sin2x\cdot Cos3x=0\\\\\\\left[\begin{array}{ccc}Sin2x=0\\Cos3x=0\end{array}\right\\\\\\\left[\begin{array}{ccc}2x=\pi n,n\in Z\\3x=\dfrac{\pi }{2} +\pi n,n\in Z\\\end{array}\right\\\\\\\left[\begin{array}{ccc}x=\dfrac{\pi n}{2} ,n\in Z\\x=\dfrac{\pi }{6} +\dfrac{\pi n}{3} ,n\in Z\end{array}\right\\\\\\Otvet \ : \ \frac{\pi n}{2} \ \ ; \ \ \frac{\pi }{6}+\frac{\pi n}{3} ,n\in Z$

$\displaystyle\bf\\2)\\\\Cos2x+5Sinx-4=0\\\\1-2Sin^{2} x+5Sinx-4=0\\\\2Sin^{2} x-5Sinx+3=0\\\\Sinx=m \ \ ; \ \ -1\leq m\leq 1\\\\2m^{2} -5m+3=0\\\\D=(-5)^{2} -4\cdot 2\cdot 3=25-24=1\\\\\\m_{1} =\frac{5-1}{4} =1\\\\\\m_{2} =\frac{5+1}{4} =1,5 > 1 \ - \ ne \ podxodit\\\\\\Sinx=1\\\\\boxed{x=\frac{\pi }{2} +2\pi n,n\in Z}$