Question

решить уравнение 8cos^2x+6cosx-5=0

Answer 1

8Cos²x + 6Cosx - 5 =0

Cosx = t ,  - 1 ≤ t ≤ 1

$8t^{2}+6t-5=0\\\\D=6^{2}-4*8*(-5)=36+160=196=14^{2}\\\\t_{1}=\frac{-6+14}{16}=\frac{1}{2}\\\\t_{2}=\frac{-6-14}{16}=-1,25<-1-neyd\\\\Cosx=\frac{1}{2} \\\\x=\pm arcCos\frac{1}{2}+2\pi n,n\in Z\\\\x=\pm\frac{\pi }{3}+2\pi n,n\in Z\\\\Otvet:\boxed{\pm\frac{\pi }{3}+2\pi n,n\in Z}$