Question

2cos^2x+cosx-1=0

допоможіть будь ласка ​

Answer 1

$\displaystyle\bf\\2Cos^{2}x+Cosx-1=0\\\\Cosx=m \ \ , \ \ -1\leq m\leq 1\\\\2m^{2} +m-1=0\\\\D=1^{2} -4\cdot 2\cdot(-1)=1+8=9=3^{2} \\\\\\m_{1} =\frac{-1+3}{4} =\frac{2}{4} =\frac{1}{2} \\\\\\m_{2} =\frac{-1-3}{4} =\frac{-4}{4}=-1\\\\\\1)\\\\Cosx=\frac{1}{2} \\\\x=\pm \ arc Cos\frac{1}{2} +2\pi n,n\in Z\\\\x=\pm \ \frac{\pi }{3} +2\pi n,n\in Z\\\\2)\\\\Cosx=-1\\\\x=\pi +2\pi n,n\in Z\\\\\\Otvet \ : \ x=\pm \ \frac{\pi }{3} +2\pi n,n\in Z \ \ , \ \ x=\pi +2\pi n,n\in Z$