SOLUTION: Cos 2x + cos x + 1= 0. Solve.

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Question 558191: Cos 2x + cos x + 1= 0. Solve.
Found 2 solutions by Alan3354, richard1234:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Cos 2x + cos x + 1= 0. Solve.
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Sub u for cos)x)
u%5E2+%2B+u+%2B+1+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B1x%2B1+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%281%29%5E2-4%2A1%2A1=-3.

The discriminant -3 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -3 is + or - sqrt%28+3%29+=+1.73205080756888.

The solution is , or
Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B1%2Ax%2B1+%29

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Sub back:
cos(x) = -1/2 +/- i*sqrt(3)/2
No real solutions.

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
I presume you mean instead of .

Using trig identities, we have



Regroup this way:



Now sin^2 - 1 is equal to -cos^2 x, so we have



If cos x = 0 then x = pi/2, 3pi/2, 5pi/2, etc. Otherwise, we may divide by cos x to obtain



We obtain x = 2pi/3, 4pi/3, 8pi/3, 10pi/3, ... or in general

or or where k is any integer.