Question

Find the number of solution of the equation in [0, 2π], tan (5π cos α) = cot (5π sin α)

• A. 22
• B. 24
• C. 28
• D. 30

Answer 1

Answer: (C)

28

Answer 2

The given equation is $\tan (5\pi \cos \alpha) = \cot (5\pi \sin \alpha)$.

Using the identity $\cot x = \tan (\frac{\pi}{2} - x)$, we can rewrite the equation as: $\tan (5\pi \cos \alpha) = \tan (\frac{\pi}{2} - 5\pi \sin \alpha)$

The general solution for $\tan A = \tan B$ is $A = n\pi + B$, where $n$ is an integer. So, $5\pi \cos \alpha = n\pi + (\frac{\pi}{2} - 5\pi \sin \alpha)$.

Dividing the entire equation by $\pi$: $5 \cos \alpha = n + \frac{1}{2} - 5 \sin \alpha$

Rearranging the terms to group $\cos \alpha$ and $\sin \alpha$: $5 \cos \alpha + 5 \sin \alpha = n + \frac{1}{2}$

Divide by 5: $\cos \alpha + \sin \alpha = \frac{n}{5} + \frac{1}{10} = \frac{2n+1}{10}$

We can express the left side, $\cos \alpha + \sin \alpha$, in the form $R \sin(\alpha + \beta)$. Using the R-factorization method, $R = \sqrt{1^2 + 1^2} = \sqrt{2}$, and $\beta = \frac{\pi}{4}$. So, $\cos \alpha + \sin \alpha = \sqrt{2} \sin(\alpha + \frac{\pi}{4})$.

Substituting this back into the equation: $\sqrt{2} \sin(\alpha + \frac{\pi}{4}) = \frac{2n+1}{10}$

$\sin(\alpha + \frac{\pi}{4}) = \frac{2n+1}{10\sqrt{2}}$

Let $X = \alpha + \frac{\pi}{4}$. Since $\alpha \in [0, 2\pi]$, the range for $X$ is $[\frac{\pi}{4}, 2\pi + \frac{\pi}{4}] = [\frac{\pi}{4}, \frac{9\pi}{4}]$. This interval has a length of $2\pi$.

The equation becomes $\sin X = C$, where $C = \frac{2n+1}{10\sqrt{2}}$. For real solutions to exist, we must have $-1 \le C \le 1$. $-1 \le \frac{2n+1}{10\sqrt{2}} \le 1$ $-10\sqrt{2} \le 2n+1 \le 10\sqrt{2}$

Since $10\sqrt{2} \approx 14.14$, the inequality becomes: $-14.14 \le 2n+1 \le 14.14$

Since $2n+1$ must be an odd integer, the possible values for $2n+1$ are: $\{-13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13\}$. This gives a total of 14 possible values for $2n+1$.

For each of these 14 values of $2n+1$, we get a corresponding value for $C$. Note that $C$ cannot be $1$ or $-1$ because $2n+1 = \pm 10\sqrt{2}$ has no integer solution for $n$. Also, $C$ cannot be $\pm \frac{1}{\sqrt{2}}$ because $\frac{2n+1}{10\sqrt{2}} = \pm \frac{1}{\sqrt{2}} \implies 2n+1 = \pm 10$, which has no integer solution for $n$. Therefore, for each of the 14 values of $n$, $C$ is strictly between $-1$ and $1$, and $C \neq \pm \frac{1}{\sqrt{2}}$.

For any value of $C$ such that $-1 < C < 1$ and $C \neq 0$, the equation $\sin X = C$ has exactly two solutions in any interval of length $2\pi$. Our interval for $X$ is $[\frac{\pi}{4}, \frac{9\pi}{4}]$, which has a length of $2\pi$. Since none of the possible values of $C$ are $0$, each of the 14 values of $n$ leads to exactly 2 solutions for $X$ in the interval $[\frac{\pi}{4}, \frac{9\pi}{4}]$.

This gives a total of $14 \times 2 = 28$ solutions for $X$.

We must also consider the domain restrictions for the original equation:

1. $\tan(5\pi \cos \alpha)$ is defined if $5\pi \cos \alpha \neq \frac{\pi}{2} + k\pi$, which simplifies to $\cos \alpha \neq \frac{1}{10} + \frac{k}{5}$ for any integer $k$.
2. $\cot(5\pi \sin \alpha)$ is defined if $5\pi \sin \alpha \neq m\pi$, which simplifies to $\sin \alpha \neq \frac{m}{5}$ for any integer $m$.

It can be shown that for the determined values of $n$, none of the solutions for $\alpha$ derived from $\sin(\alpha + \frac{\pi}{4}) = \frac{2n+1}{10\sqrt{2}}$ will lead to $\cos \alpha = \frac{1+2k}{10}$ or $\sin \alpha = \frac{m}{5}$. Therefore, all 28 solutions are valid.

The number of solutions is 28.