Question

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

Answer 1

@countonme123

Answer 2

you solve it like quadratic :
let cosx = y
then
2y^2 + 3y + 1 =0
2y^2 +2y + y + 1 =0
2y(y+1) + 1(y+1)=0
(2y+1)(y+1) =0

y = -1 , y = -1/2
so
cos(x) = -1 , cos(x) = -1/2

can you answer now ?

Answer 3

(another thing to remember is that those functions are periodic so for cos(x) = -1/2 there are some solutions for example)

Answer 4

So it is c right?

Answer 5

b.... nvm

Answer 6

b