Find all values of the parameter a for which the equation cos(sqrt(a - x^2)) = 1 has exactly 8 roots.

Step-by-step solution

1. 1

   Fix the domain first. The square root demands $a-x^{2}\geq 0$, i.e. $x^{2}\leq a$. If $a<0$ this is impossible for every real $x$, so there are no roots at all and we may assume

   $a\geq 0$

2. 2

   Convert the cosine equation into a family of equations. Since $\cos t=1$ exactly when $t=2\pi n$ for integer $n$, and since $t=\sqrt{a-x^{2}}\geq 0$ forces $2\pi n\geq 0$, only $n=0,1,2,\dots$ can occur:

   $\sqrt{a-x^{2}}=2\pi n\quad\Longrightarrow\quad x^{2}=a-4\pi^{2}n^{2}$

   Different $n$ give different values of $4\pi^{2}n^{2}$, so the root sets for different $n$ never overlap — no double counting.

3. 3

   Count how many roots each $n$ contributes. For a fixed admissible $n$:

   $a-4\pi^{2}n^{2}>0\ \Rightarrow\ x=\pm\sqrt{a-4\pi^{2}n^{2}}\ \ (\text{two roots}),\qquad a-4\pi^{2}n^{2}=0\ \Rightarrow\ x=0\ \ (\text{one root})$

   and $a-4\pi^{2}n^{2}<0$ contributes nothing. So the total is $2k$ or $2k+1$, where $k=\#\{n\geq 0:4\pi^{2}n^{2}<a\}$, the $+1$ appearing precisely when $a$ equals some $4\pi^{2}n^{2}$.

4. 4

   Rule out the exceptional values of $a$. If $a=4\pi^{2}m^{2}$ for some integer $m\geq 0$, then the strict inequality $4\pi^{2}n^{2}<a$ holds for $n=0,1,\dots,m-1$, so $k=m$ and the total root count is $2m+1$ — always odd, hence never $8$. So $a$ is not of that form.

5. 5

   Solve $2k=8$ for the range of $a$. We need $k=4$, i.e. exactly the four values $n=0,1,2,3$ satisfy $4\pi^{2}n^{2}<a$ while $n=4$ does not:

   $4\pi^{2}\cdot 3^{2}<a\leq 4\pi^{2}\cdot 4^{2}\ \Longrightarrow\ 36\pi^{2}<a\leq 64\pi^{2}$

   The right endpoint $a=64\pi^{2}$ is one of the excluded exceptional values (it would give $2\cdot 4+1=9$ roots), so it must be dropped:

   $36\pi^{2}<a<64\pi^{2}$

6. 6

   Verify the boundaries numerically. Counting roots by the formula above: $a=36\pi^{2}\approx 355.306$ gives $7$ roots, $a=36\pi^{2}+0.01$ gives $8$, $a=50\pi^{2}\approx 493.480$ gives $8$, $a=64\pi^{2}-0.01$ gives $8$, and $a=64\pi^{2}\approx 631.655$ gives $9$. For $a=50\pi^{2}$ the eight explicit roots are $\pm 11.7548,\pm 18.3185,\pm 21.3073,\pm 22.2144$, and substituting each back gives $\cos\sqrt{a-x^{2}}=1$ to within $10^{-9}$.