Find all values of the parameter a for which the equation cosine of the square root of (a minus x squared) equals 1 has exactly 8 roots.

Step-by-step solution

1. 1

   Translate the cosine equation into a list of levels. $\cos\theta=1$ exactly when $\theta=2\pi k$ for an integer $k$, and here $\theta=\sqrt{a-x^2}\ge0$, so only $k=0,1,2,\dots$ are possible:

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

   The equation has become a family of simple quadratics, one per $k$.

2. 2

   Count how many roots each level contributes. For a fixed $k$:

   • if $a-4\pi^2k^2>0$ there are two roots, $x=\pm\sqrt{a-4\pi^2k^2}$;
   • if $a-4\pi^2k^2=0$ there is exactly one root, $x=0$;
   • if $a-4\pi^2k^2<0$ there are none.

   Different $k$ give different values of $|x|$, so no root is ever double-counted.

3. 3

   Rule out the boundary cases immediately. If $a=4\pi^2m^2$ for some integer $m$, the total is $2m+1$, an odd number. Since $8$ is even, no such $a$ can work, and every level must be strictly cleared.

4. 4

   Require exactly four usable values of $k$. With no equalities, the count is $2\times(\text{number of }k\text{ with }4\pi^2k^2<a)$. Setting this to $8$ means $k=0,1,2,3$ must all satisfy the strict inequality while $k=4$ must fail:

   $4\pi^2\cdot3^2<a\le 4\pi^2\cdot4^2$

5. 5

   Trim the closed endpoint. The right end $a=64\pi^2$ is one of the excluded boundary values (it gives $2\cdot4+1=9$ roots), so it must be dropped, and $a=36\pi^2$ gives $2\cdot3+1=7$. The answer is the open interval

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

6. 6

   Check the ends by counting. At $a=50\pi^2$ (inside the interval) the levels $k=0,1,2,3$ all clear and $k=4$ does not, giving $2\cdot4=8$ roots. At $a=36\pi^2$ the count drops to $7$ and at $a=64\pi^2$ it jumps to $9$, so the strict inequalities on both ends are essential.