SOLUTION: Find the exact solutions of the given equation in the interval [0,2pi).
cos2x+3cosx+2=0
a)x=2pi/3,pi,4pi/3
b)x=0,pi/3,2pi/3,pi,4pi/3
c)x=pi/2,7pi/6,11pi/6
d)x=0
e)x=0,pi/3,
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Trigonometry-basics
-> SOLUTION: Find the exact solutions of the given equation in the interval [0,2pi).
cos2x+3cosx+2=0
a)x=2pi/3,pi,4pi/3
b)x=0,pi/3,2pi/3,pi,4pi/3
c)x=pi/2,7pi/6,11pi/6
d)x=0
e)x=0,pi/3,
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Question 937541: Find the exact solutions of the given equation in the interval [0,2pi).
cos2x+3cosx+2=0
a)x=2pi/3,pi,4pi/3
b)x=0,pi/3,2pi/3,pi,4pi/3
c)x=pi/2,7pi/6,11pi/6
d)x=0
e)x=0,pi/3,pi,5pi/3 Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! cos2x+3cosx+2=0
2cos^2x - 1 + 3cosx + 2 = 0
2cos^2x + 3cosx + 1 = 0
(2cosx + 1)(cosx + 1) = 0
cosx = -1/2 0r cosx = -1
....
a)x=2pi/3 ,pi, 4pi/3
........
(cosx, sinx) Summary Unit Circle .
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