Question

The number of solutions of the equation $\sec x \cos 5x + 1 = 0$ in the interval $[0, 2\pi]$ is

• A. \( 5 \)
• B. \( 8 \)
• C. \( 10 \)
• D. \( 12 \)

Hint:

When solving trigonometric equations involving secant, cosecant, tangent, or cotangent, always remember to consider the domain restrictions. For $\sec x = \frac{1}{\cos x}$, the restriction is $\cos x \neq 0$. After finding the general solutions, filter them based on the given interval and exclude any values that make the original expression undefined. For equations like $\cos A = -\cos B$, rewrite $-\cos B$ as $\cos(\pi \pm B)$ to use the general solution formula for cosine.

Answer 1

The Correct Option is B

The given equation is $\sec x \cos 5x + 1 = 0$. First, rewrite $\sec x$ as $\frac{1}{\cos x}$: $\frac{\cos 5x}{\cos x} + 1 = 0$ For the expression to be defined, $\cos x \neq 0$. This means $x \neq \frac{\pi}{2}$ and $x \neq \frac{3\pi}{2}$ in the interval $[0, 2\pi]$. Rearrange the equation: $\cos 5x = -\cos x$ We know that $-\cos x = \cos(\pi - x)$ or $\cos(\pi + x)$. We use the general solution for $\cos A = \cos B$, which is $A = 2n\pi \pm B$, where $n$ is an integer. Case 1: $\cos 5x = \cos(\pi - x)$ $5x = 2n\pi \pm (\pi - x)$ Subcase 1.1: $5x = 2n\pi + (\pi - x)$ $5x = 2n\pi + \pi - x$ $6x = (2n+1)\pi$ $x = \frac{(2n+1)\pi}{6}$ For $x \in [0, 2\pi]$, we find the values of $n$: For $n=0, x = \frac{\pi}{6}$ For $n=1, x = \frac{3\pi}{6} = \frac{\pi}{2}$ For $n=2, x = \frac{5\pi}{6}$ For $n=3, x = \frac{7\pi}{6}$ For $n=4, x = \frac{9\pi}{6} = \frac{3\pi}{2}$ For $n=5, x = \frac{11\pi}{6}$ (For $n=6, x = \frac{13\pi}{6}$, which is outside the interval) Subcase 1.2: $5x = 2n\pi - (\pi - x)$ $5x = 2n\pi - \pi + x$ $4x = (2n-1)\pi$ $x = \frac{(2n-1)\pi}{4}$ For $x \in [0, 2\pi]$, we find the values of $n$: For $n=1, x = \frac{\pi}{4}$ For $n=2, x = \frac{3\pi}{4}$ For $n=3, x = \frac{5\pi}{4}$ For $n=4, x = \frac{7\pi}{4}$ (For $n=5, x = \frac{9\pi}{4}$, which is outside the interval) Combining all solutions obtained: $S_1 = \left\{ \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{3\pi}{2}, \frac{11\pi}{6} \right\}$ $S_2 = \left\{ \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \right\}$ Now, we must consider the restriction $\cos x \neq 0$. This means we must exclude $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ from our set of solutions. The solutions $\frac{\pi}{2}$ and $\frac{3\pi}{2}$ are present in $S_1$. Removing the excluded values: From $S_1$: $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ (4 solutions) From $S_2$: $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ (4 solutions) All these solutions are distinct. The total number of solutions in the interval $[0, 2\pi]$ is $4 + 4 = 8$. The final answer is $\boxed{8}$.