Question

The number of distinct real roots of \[ \begin{vmatrix} \sin x & \cos x \\ \cos x & \sin x \end{vmatrix} = 0 \] in the interval \[ -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} \] is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Always pay close attention to whether the interval boundaries are inclusive (closed interval $[\dots]$) or exclusive (open interval $(\dots)$).
Here, the closed interval includes the boundary values $-\pi/4$ and $\pi/4$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We first expand the given determinant and then solve the resulting trigonometric equation within the specified closed interval.
Key Formula or Approach:
The determinant of a $2 \times 2$ matrix is given by:
\[ \begin{vmatrix} a & b c & d \end{vmatrix} = ad - bc \]
We also use the double-angle trigonometric identity:
\[ \cos 2x = \cos^2 x - \sin^2 x \]

Step 2: Detailed Explanation:
Expand the given determinant:
\[ \begin{vmatrix} \sin x & \cos x \cos x & \sin x \end{vmatrix} = 0 \implies \sin^2 x - \cos^2 x = 0 \]
Multiply the entire equation by $-1$:
\[ \cos^2 x - \sin^2 x = 0 \implies \cos 2x = 0 \]
Now, let us find the solutions for $x$ in the interval $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$.
If $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$, then the range for $2x$ is:
\[ -\frac{\pi}{2} \le 2x \le \frac{\pi}{2} \]
Within this interval, $\cos 2x = 0$ occurs at:
\[ 2x = -\frac{\pi}{2} \quad \text{and} \quad 2x = \frac{\pi}{2} \]
Divide by 2 to find the values of $x$:
\[ x = -\frac{\pi}{4} \quad \text{and} \quad x = \frac{\pi}{4} \]
Both values lie within the closed interval $-\frac{\pi}{4} \le x \le \frac{\pi}{4}$.
Therefore, there are exactly 2 distinct real roots.

Step 3: Final Answer
The correct option is (C).