Finding parameter values from a sine equation with radical domain

Key takeaway: The equation sin(sqrt(a - x^2)) = 0 combines a trigonometric zero condition with a radical domain restriction, so the solution set must be built carefully before summing values.

Step 1: Use the zero condition of sine

For sin(sqrt(a - x^2)) = 0, the inside angle must be an integer multiple of pi:

$\sqrt{a-x^2}=k\pi, \quad k\in\mathbb{Z}$

Because the square root is always nonnegative, we only need k \ge 0. Squaring gives

$a-x^2=k^2\pi^2 \quad \Rightarrow \quad x^2=a-k^2\pi^2$

So each admissible k produces solutions

$x=\pm\sqrt{a-k^2\pi^2}$

provided $a-k^2\pi^2\ge 0$.

Step 2: Describe the full solution set

The allowed integers k are those with

$k^2\pi^2\le a$

If $a<0$, there are no solutions. If $a\ge 0$, then each positive k contributes a symmetric pair $\pm x$, whose sum is 0. Only the k = 0 case can contribute a single repeated value:

$\sqrt{a-x^2}=0 \Rightarrow x^2=a \Rightarrow x=\pm\sqrt{a}$

That pair also sums to 0. Therefore every valid solution set is symmetric about 0, so the sum of all solutions is always 0 whenever solutions exist.

Step 3: Compare with the required sum

The problem asks for the sum of all solutions to be 100. But the symmetry of the equation forces that sum to be 0, not 100.

So there is no real value of $a$ for which the condition can hold.

Common structural reason

The expression depends on $x^2$ only, so the equation is even in $x$. Any solution $x$ implies $-x$ is also a solution. That pairing makes the total sum cancel unless there is an unpaired solution, which cannot happen here because the equation still depends on $x^2$.

Hence the set of parameter values is empty.

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Pitfalls the pros know 👇 A frequent mistake with sin(sqrt(a - x^2)) = 0 is to focus only on the equation inside the square root and forget the symmetry in x. If you write the solutions as x = plus or minus something, the pair always cancels in the sum. Another trap is to assume the integer parameter k can be negative and create extra distinct roots, but the square root output cannot be negative, so only nonnegative k matter. It is also easy to overlook that even the k = 0 case gives ±sqrt(a), which still sums to zero. Because of that cancellation, any target sum other than zero is impossible.

What if the problem changes? If the problem were changed to find all values of a for which the sum of all solutions of sin(sqrt(a - x^2)) = 0 is equal to 0, then every real a that allows at least one solution would work. For example, a = 0 gives only x = 0, and a = 10 gives the symmetric pair x = ±sqrt(10) from the k = 0 case, along with any additional symmetric pairs from larger k values if they fit the domain. The key change is the required sum: zero is compatible with the even symmetry of the equation, while 100 is not. If the parameter target became 2 instead of 100, the answer would still be no real value of a, because the solution set remains symmetric for every admissible a.

Tags: zeroes of sine function, radical equation domain, symmetric solution pairs