<s>[INST] Values for $a$ and $b$ in $y=\cos(x)+a\cos(bx)$ such that every real value for $x$ has either a positive or $0$ value for $y$ If there is a function in the form $y=\cos(x)+a\cos(bx)$ do there exists real number values for $a$ and $b$ such that for every real number value for $x$ there is either a positive number value for $y$ or a $0$ value for $y$?
 [/INST] Let $f(x) = \cos(x)+a\cos(bx)$, and suppose $f(x) \ge 0$ for all $x \in \mathbb{R}$.

First suppose $b=0$. Then $f(x) = \cos(x) + a,$ which has range $[a-1,\infty).$

Thus, for $b=0$, we get the solution pairs $(a,b) = (t,0),\,$ for any $t \ge 1.$

Next, suppose $b \ne 0$.

If $|a| > 1$, then when $a\cos(bx)$ realizes a value of $-a$, $f(x)$ will be negative, contradiction.

Similarly, if $|a| < 1$, then when $\cos(x)$ realizes a value of $-1$, $f(x)$ will be negative, contradiction.

It follows that $|a| = 1,$ hence either
\begin{align*}
f(x) &= \cos(x) + \cos(bx)\\[4pt]
&\,\text{or}\\[4pt]
f(x) &= \cos(x) - \cos(bx)\\[4pt]
\end{align*}
First suppose $f(x) = \cos(x) + \cos(bx)$. 

Then when one of the summands realizes a value of $-1$, the other must realize a value of $+1$ to compensate. But that means each of the summands has a period which is a multiple of the other. It follows that periods are equal, hence $b = \pm 1$. But $\cos\,$ is an even function hence, $b = \pm 1$ implies $\cos(bx) = \cos(x)$. But then $f(x) = 2\cos(x)$, contradiction, since then $f(\pi) < 0$.

Next suppose $f(x) = \cos(x) - \cos(bx)$. 

If $x_0$ is such that $\cos(x_0) = -1$, then $\cos(bx_0)$ must also be $-1$, else $f(x_0) < 0$.

It follows that the period of $\cos(bx)$ divides the period of $\cos(x)$.

Similarly, if $x_0$ is such that $\cos(bx_0) = 1$, $\cos(x_0)$ must also be $1$, else $f(x_0) < 0$.

It follows that the period of $\cos(x)$ divides the period of $\cos(bx)$.

Therefore the periods must be equal, hence $b = \pm 1$. But then, since $\cos$ is an even function, $b = \pm 1$ implies $\cos(bx) = \cos(x)$, hence $f(x)$ is identically $0$.

This gives the solution pairs $(a,b) = (-1,\pm 1).$

To summarize, we have the solution pairs
\begin{align*}
(a,b) &= (-1,1)\\[4pt]
(a,b) &= (-1,-1)\\[4pt]
(a,b) &= (t,0),\,\text{for any $t \ge 1$}\\[4pt]
\end{align*}
and those are the only solutions.
</s>