Answer
See explanations.
Work Step by Step
Step 1. To show that the function $r(\theta)=2\theta-cos^2(\theta)+\sqrt 2$ has exactly one zero in the interval of $(-\infty,\infty)$, we will need to show that the function crosses the horizontal axis only once. (Please note that this equation is not in polar form.)
Step 2. Taking the derivative of the function, we get $r'(\theta)=2+2sin(\theta) cos(\theta)=2+sin(2\theta)$. As $|sin(2\theta)|\leq1$, we have $r'(\theta)\gt0$, which means that the function $r(\theta)$ will only go in one direction (increases as $\theta$ increases).
Step 3. Let $\theta=-\pi/2$; we have $r(-\pi/2)=\sqrt 2-\pi\lt0$, which is negative. Let $\theta=\pi/2$; we have $r(\pi/2)=\sqrt 2+\pi\gt0$, thus the curve should cross the horizontal axis between $-\pi/2$ and $\pi/2$. As a matter of fact, we can find the zero as $\theta=-0.234$ as shown in the figure.
Step 4. Because $r'(\theta)\gt0$, the function has only one solution at $\theta=-0.234$.
