Prove that ()+2x +3=0 has at least one real root.
prove that x=cosx has at least one solution.

The first one is of order 5, so it has either 1, 3 or 5 real roots (unless any coefficent was complex). Proof complete :)
The other one, if it has a solution, it must be in [-1;1]. Because it only gives positive results the solution is further restricted to [0;1]. Because the cosine function is continuous and strictly decreasing on this interval, the difference of x and it's cosine will shrink up to some point within the interval where it gets to 0 (the solution) and then flips sign (the cosine gets less than the number), further decreasing until the end of the interval.

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