Question

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

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The Correct Option is C

To find the number of distinct real roots of the equation $\begin{vmatrix}\cos x & \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x\end{vmatrix}= 0$ in the interval $\left[- \frac{\pi}{4}, \frac{\pi}{4}\right]$, we need to evaluate the determinant. A square matrix determinant reads zero when rows or columns are linearly dependent.

The given matrix is a 3x3 symmetric matrix. Below is the expanded form of its determinant:

$ \begin{vmatrix} \cos x & \sin x & \sin x\\ \sin x & \cos x & \sin x\\ \sin x & \sin x & \cos x \end{vmatrix} $
This can be expanded as:
$\cos x[(\cos x)(\cos x) - (\sin x)(\sin x)]$ -$\sin x[(\sin x)(\sin x) - (\cos x)(\sin x)]$ +$\sin x[(\sin x)(\sin x) - (\cos x)(\sin x)]$

Simplifying, we get:

$ = \cos x(\cos^2 x - \sin^2 x) - \sin x(\sin^2 x - \cos x \sin x) + \sin x(\sin^2 x - \cos x \sin x) $

Using the trigonometric identity $\cos^2 x - \sin^2 x = \cos 2x$, the determinant simplifies to:

$ = \cos x \cdot \cos 2x - \sin x \cdot (0) + \sin x \cdot (0) $

Thus, the determinant becomes:

$ = \cos x \cdot \cos 2x $

The equation $\cos x \cdot \cos 2x = 0$ leads to two separate equations:

1. $\cos x = 0$
2. $\cos 2x = 0$

Let's solve both equations in the given interval $\left[- \frac{\pi}{4}, \frac{\pi}{4}\right]$:

1. Solving $\cos x = 0$

In the interval $\left[- \frac{\pi}{4}, \frac{\pi}{4}\right]$, $\cos x$ cannot be zero because $x = \pm \frac{\pi}{2}$ is outside this range.

2. Solving $\cos 2x = 0$

This implies:

$ 2x = \pm \frac{\pi}{2} $ which simplifies to

$ x = \pm \frac{\pi}{4} $

Both solutions, $x = \frac{\pi}{4}$ and $x = -\frac{\pi}{4}$, lie within the given interval.

Conclusion

Hence, there are 2 distinct real roots of the equation in the interval $\left[- \frac{\pi}{4}, \frac{\pi}{4}\right]$.