Question

2cos2x+1=0 решите плз, за спам взломаю акк

Answer 1

$2\, cos2x+1=0\\\\cos2x=-\dfrac{1}{2}\\\\2x=\pm arccos(-\dfrac{1}{2})+2\pi n\ ,\ n\in Z\\\\2x=\pm (\pi -arccos\dfrac{1}{2})+2\pi n\ ,\ n\ib Z\\\\2x=\pm (\pi -\dfrac{\pi }{3})+2\pi n\ ,\ n\in Z\\\\2x=\pm \dfrac{2\pi }{3}+2\pi n\ ,\ n\in Z\\\\\boxed{\ x=\pm \dfrac{\pi}{3}+\pi n\ ,\ n\in Z\ }$

Answer 2

cos2x=-1/2

2x=+-arccos (-1/2)+2πk     k∈Z

Т.к. arccos(-a)+arcos (a)=π  ⇒arccos(-1/2)=π-arccos(1/2)=π-π/3=2/3π

2x=+-2π/3+2πk        x=+-(π/3)+πk    k∈Z