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# cos2x+5cosx+3=0

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cos2x+5cosx+3=0

Guest Aug 21, 2015

#1
+80983
+10

cos2x+5cosx+3=0   ....note that cos2x = 2cos^x - 1  ....so we have.......

2cos^x - 1 + 5 cosx + 3  = 0     simplify

2cos^x + 5cosx + 2   = 0      factor

(2cosx + 1) (cosx + 2)  = 0     setting each factor to 0, we have......

2cosx + 1 = 0                             and                           cosx + 2 = 0

subtract 1 from each side                                            subtract 2 from each side

2cosx = -1                                                                cos x = -2

divide both sides by 2                                                this has no solution

cosx = -1/2

And the solution to this is  x  = (2/3)pi + 2pi * n    and     x = (4/3)pi + 2pi * n ....   where n is any integer

Here's a graph......https://www.desmos.com/calculator/wb2x5nbzfv

CPhill  Aug 21, 2015
Sort:

#1
+80983
+10

cos2x+5cosx+3=0   ....note that cos2x = 2cos^x - 1  ....so we have.......

2cos^x - 1 + 5 cosx + 3  = 0     simplify

2cos^x + 5cosx + 2   = 0      factor

(2cosx + 1) (cosx + 2)  = 0     setting each factor to 0, we have......

2cosx + 1 = 0                             and                           cosx + 2 = 0

subtract 1 from each side                                            subtract 2 from each side

2cosx = -1                                                                cos x = -2

divide both sides by 2                                                this has no solution

cosx = -1/2

And the solution to this is  x  = (2/3)pi + 2pi * n    and     x = (4/3)pi + 2pi * n ....   where n is any integer

Here's a graph......https://www.desmos.com/calculator/wb2x5nbzfv

CPhill  Aug 21, 2015

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