solve please?

Question

solve the equation from 0,2pie

cos^2x+2cosx+1

please explain thank you

Answer 1

I don't think you write the equation in a right way.
Maybe cos(2x)+2cosx+1=0 is what you want

Answer 2

I think the teacher meant you to calculate the value of cos^2x + 2cosx + 1 for different vaues of x between 0 and 2pi.

x ----- Value of expression

0 ------ 1 + 2 + 1 = 4 since cos 0 = 1

pi/2 ------ 0 + 0 + 1 = 1 since cos pi/2 = cos 90 = 0

pi ------- 1 - 2 + 1 = 0 since cos pi = cos 180 = -1

270 ------- 0 + 0 + 1 = 1 since cos 3pi/2 = cos 270 = 0

360 ------- 1 + 2 + 1 = 4 since cos 2pi = cos 360 = 1

Answer 3

I think you're all screwed up on the equation ☺
Actually, since you don't have an equals sign, there --is-- no equation. And, if you meant
cos^2x+2cosx+1 = 0 then you'll have an equation with complex-valued roots and I have the sense that you're probably not quite ready for complex analysis yet. (If you were, you wouldn't be having problems with this little equation ☺)

Doug

Answer 4

Doug, you're wrong.
If the equation is indeed cos^2x+2cosx+1=0 (that is,
(cosx)^2+2 cos x +1 = 0,
then
cos x = -1
and x = (2k+1) * PI,
k integer

Answer 5

cos^2x+2cosx+1= 0 <-- I assume you meant this
(cos x +1)^2=0
cosx+1 = 0 <-- Take square root of both sides
cos x=-1
x = pi = 180 degrees

Answer 6

I'd punch it into a calculator and use the graph to find when it = 0. I'm not sure there is even a way to do that analytically.