Proving $x^5+\frac{1}{|\cos x|+1}=1$ has at least one solution in $(0,\infty)$ 
Prove that the equation $$x^5+\frac{1}{|\cos x|+1}=1$$ has at least one solution in $(0,\infty)$

Does anyone know how to solve this? Thanks in advance.
 A: Let $$f (x)=x^5+\frac {1}{|\cos (x)|+1}-1$$
$f $ is continuous at $[0,\pi/2] $,
$$f (0)=-1/2 <0$$
and
$$f (\pi/2)>0$$
thus by IVT, the equation $f (x)=0$ has at least a solution at $(0,\pi/2) $.
A: For a solution to exist in (0, $\infty$), $x$ must be less than 1, for we know that the $x^5$ is greater than 1 for $x>1$ and further, that the value of the expression $\frac{1}{|cos(x)| + 1}$ is always positive for $x$ $\epsilon$ $(0,1)$.
Now, we can easily deduce that $x^5$ is strictly increasing, continuous and differentiable, in $(0,1)$.
For the expression, $\frac{1}{|cos(x)| + 1}$, we know from properties of $cos$ that it is positive, strictly decreasing, continuous and differentiable in $(0, \pi/2)$ and therefore in $(0,1)$ too. $cos(x)$ being positive in $(0,1)$, we can drop the absolute value operator around it. Now $1+cos(x)$ is also decreasing in $(0,1)$. But once we invert $1+cos(x)$ (which is $\frac{1}{1+|cos(x)|}$), we get a strictly increasing function which ranges from $1/(1+cos(0))$ i.e. $1/2$, to, $1/(1+cos(1))$, i.e. 0.649.. in the interval $(0,1)$.
Now we have 2 expressions, one ($x^5$) which ranges from $0$ to $1$ and the other $\frac{1}{1+|cos(x)|}$ which ranges from $0.5$ to $0.649..$, both in the interval of (0,1).
We can conclusively say that sum of the both the expressions must range from $(0.5, 1.649)$. Also, we already know that both expressions are continuous and differentiable in $(0,1)$, so there must be a value of x $\epsilon$ $(0,1)$ such the sum of both expressions is 1, which can be one solution.
