Question

Solve on the interval [0,2pi) 2cos^2x+3cos x+1=0

Answer 1

π, 2π/3, 4π/3

Explanation:

2cos²(x)+3cos(x)+1=0

1) Lets give cos x=y

So, we can rewrite our equation

2y²+3y+1=0

(2y+1)(y+1)=0

2y+1=0 or y+1=0

2y+1=0, 2y= -1 , y = -1/2

y+1=0, y= - 1

2) cos(x)= - 1, x=π

cos(x)= - 1/2, x=2π/3 or 4π/3