Question

$\sin fx = 8\sin{\sqrt{x+T}}$ $\sqrt{x+T} = n\pi + (-1)^n f$

• A. The condition for the existence of solutions $(x, T)$ is $|\sin(f)| \le 1/8$.
• B. The condition for the existence of solutions $(x, T)$ is $|\sin(f)| \ge 1/8$.
• C. The condition for the existence of solutions $(x, T)$ is $|\cos(f)| \le 1/8$.
• D. The condition for the existence of solutions $(x, T)$ is $|\cos(f)| \ge 1/8$.

Answer 1

Answer: (A)

The condition for the existence of solutions $(x, T)$ is $|\sin(f)| \le 1/8$.

Answer 2

Let $\theta = \sqrt{x+T}$. From the second equation, $\theta = n\pi + (-1)^n f$. Substituting into the first equation gives $\sin(fx) = 8\sin(\theta) = 8\sin(n\pi + (-1)^n f)$. Using the identity $\sin(n\pi + \phi) = (-1)^n \sin(\phi)$ is not directly applicable here in the way it might seem. The correct evaluation of $\sin(n\pi + (-1)^n f)$ for integer $n$ simplifies to $\sin(f)$.

Case 1: $n$ is even. Let $n=2k$ for some integer $k$. Then $\sin(2k\pi + (-1)^{2k} f) = \sin(2k\pi + f) = \sin(f)$.

Case 2: $n$ is odd. Let $n=2k+1$ for some integer $k$. Then $\sin((2k+1)\pi + (-1)^{2k+1} f) = \sin((2k+1)\pi - f) = \sin(\pi - f) = \sin(f)$.

In both cases, $\sin(n\pi + (-1)^n f) = \sin(f)$.

So, the first equation becomes $\sin(fx) = 8\sin(f)$. For this equation to have a real solution for $x$, the value $8\sin(f)$ must lie within the range of the sine function, which is $[-1, 1]$. Therefore, we must have: $|8\sin(f)| \le 1$ $|\sin(f)| \le \frac{1}{8}$

This condition is necessary for the existence of real solutions for $x$ and $T$. To show it is sufficient, assume $|\sin(f)| \le 1/8$. If $f=0$, the condition is satisfied. The equations become $\sin(0) = 8\sin(\sqrt{x+T})$ and $\sqrt{x+T} = n\pi$. This simplifies to $0 = 8\sin(n\pi)$ and $\sqrt{x+T} = n\pi$. $0=0$ is always true. We need $\sqrt{x+T} = n\pi$. For $n=1$, $\sqrt{x+T} = \pi$, so $x+T = \pi^2$. This can be satisfied, e.g., $x=0, T=\pi^2$. If $f \ne 0$, let $Y = 8\sin(f)$. Since $|Y| \le 1$, $\sin(fx) = Y$ has real solutions for $x$. For example, $fx = \arcsin(Y)$, so $x = \frac{1}{f}\arcsin(8\sin(f))$. From the second equation with $n=0$, we have $\sqrt{x+T} = f$. Squaring gives $x+T = f^2$. Then $T = f^2 - x = f^2 - \frac{1}{f}\arcsin(8\sin(f))$. Thus, real solutions for $x$ and $T$ exist if and only if $|\sin(f)| \le 1/8$.