Question

Number of values of \( \theta \) lying in the interval \( (-\pi, \pi) \) such that \( \begin{vmatrix} \cos\theta & -\sin\theta & 1 \sin\theta & 1 & -\cos\theta 1 & \cos\theta & \sin\theta \end{vmatrix} = 2 \) is:

• A. 1
• B. 0
• C. 2
• D. 3

Hint:

Whenever you encounter the sum \( \sin^3\theta + \cos^3\theta \), substituting \( s = \sin\theta + \cos\theta \) often converts the trigonometric equation into a polynomial in \( s \). Always verify if the resulting roots fall within the range \( [-\sqrt{2}, \sqrt{2}] \).

Answer 1

The Correct Option is B

Concept:& nbsp;

The problem requires evaluating a \(3 \times 3\) determinant and solving the resulting trigonometric equation within the given interval.

• Determinant expansion: \[ \Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}. \]

• Range of \(\sin\theta+\cos\theta\): The expression \[ a\sin\theta+b\cos\theta \] has the range \[ \left[-\sqrt{a^2+b^2},\ \sqrt{a^2+b^2}\right]. \]

Step 1: Expand the determinant.

Let \[ \Delta= \begin{vmatrix} \cos\theta & amp; -\sin\theta & amp; 1\\ \sin\theta & amp; 1 & amp; -\cos\theta\\ 1 & amp; \cos\theta & amp; \sin\theta \end{vmatrix}. \]

Expanding along the first row,

\[ \Delta = \cos\theta(\sin\theta+\cos^2\theta) -(-\sin\theta)(\sin^2\theta+\cos\theta) +(\sin\theta\cos\theta-1). \] \[ \Delta = \sin\theta\cos\theta +\cos^3\theta +\sin^3\theta +\sin\theta\cos\theta +\sin\theta\cos\theta -1. \] \[ \boxed{\Delta=\sin^3\theta+\cos^3\theta+3\sin\theta\cos\theta-1.} \]

Step 2: Solve the equation \(\Delta=2\).

Given \[ \Delta=2, \] we get

\[ \sin^3\theta+\cos^3\theta+3\sin\theta\cos\theta=3. \]

Let \[ s=\sin\theta+\cos\theta. \] Then \[ s^2=1+2\sin\theta\cos\theta, \] so \[ \sin\theta\cos\theta=\frac{s^2-1}{2}. \]

Using the identity \[ a^3+b^3=(a+b)^3-3ab(a+b), \] we obtain

\[ s^3 -3\left(\frac{s^2-1}{2}\right)s +3\left(\frac{s^2-1}{2}\right) =3. \]

Multiplying by \(2\),

\[ 2s^3-3s^3+3s+3s^2-3=6. \] \[ -s^3+3s^2+3s-9=0. \] \[ s^3-3s^2-3s+9=0. \] \[ s^2(s-3)-3(s-3)=0. \] \[ (s^2-3)(s-3)=0. \]

Hence, the possible values of \(s\) are

\[ s=3,\quad \sqrt{3},\quad -\sqrt{3}. \]

Step 3: Check the validity of the solutions.

Since \[ s=\sin\theta+\cos\theta =\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right), \] its range is

\[ [-\sqrt{2},\ \sqrt{2}] \approx [-1.414,\ 1.414]. \]

Therefore,

• \(s=3\) is not possible.
• \(s=\sqrt{3}\approx1.732\) is not possible.
• \(s=-\sqrt{3}\approx-1.732\) is not possible.

Hence, there is no value of \(\theta\) satisfying the given equation.