Question

am i correct in thinking the only solution to

cos(x) - 1 + ((x^2)/2) = 0

is x = 0 ?

Answer 1

only real*

Answer 2

Maybe take the derivative to get a better idea?

Answer 3

well here is my argument, i dont know whether it's totally valid:

consider f(x) = sinx - x

f'(x) = cosx -1 =< 0 therefore f(x) is decreasing, since f(0) = 0 we have
sinx - x <0 for x<0
ie: sinx< x

now consider sinu < u
sin^2 (u) < u^2
-2sin^2 (u) < -2u^2
1- 2sin^2 (u) < 1 - 2u^2
cos(2u) < 1 - 2u^2

let u = x/2

now cos(x) < 1 - (x^2)/2 for 0<x

since cos(x) and 1 - (x^2)/2 are even we can say the same for x<0

therefore the only real solution for x is x = 0

Answer 4

Your logic seems correct to me. I was thinking something like:
\[f(x) =cos(x) - 1 + ((x^2)/2)\]
\[f(0) = 0 \\ f'(x) = -sin(x) + x \\ f'(0) = 0 \\ f''(x) = -cos(x) + 1 \\cos(x) \leq 1 \Rightarrow f''(x) \geq 0 \Rightarrow f'(x) \geq 0 \Rightarrow f(x) \geq 0 \]
And thus, 0 is the only solution to the original equation. Although, I think maybe your way is just as good if not better.