2cos^2 x + 3cosx + 1 = 0
Solve on the interval {0,2Pi}
A. x= Pi/6 , x = 7pi/6
B. x= Pi, x = 2pi/3, x = 4pi/3
C. x = 2pi, x = pi/3
D. x = 2pi, x = pi/4, x= 5pi/4
We will use the substitution: u = cos x 2 u² + 3 u + 1 = 0 2 u² + u + 2 u + 1 = 0 u ( 2 u + 1 ) + ( 2 u + 1 ) = 0 ( 2 u + 1 ) ( u + 1 ) = 0 2 u + 1 = 0, u = - 1/2 or: u + 1 = 0, u = - 1 cos x = - 1/2, x 1 = 2π/3, x 2 = 4π/3 cos x = - 1 , x 3 = π Answer: B )
