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If 0<x<π0 < x < \pi and the matrix [4sin⁡x−1−3sin⁡x]\begin{bmatrix} 4\sin x & -1 \\ -3 & \sin x \end{bmatrix} is singular, then the values of xx are :

Solution

✅ Correct Option: 1

Singular ⇒\Rightarrow determinant =0= 0.

4sin⁡2x−3=0⇒sin⁡2x=34⇒sin⁡x=324\sin^2 x - 3 = 0 \Rightarrow \sin^2 x = \frac{3}{4} \Rightarrow \sin x = \frac{\sqrt{3}}{2} (since sin⁡x>0\sin x > 0 on (0,π)(0,\pi)).

So x=π3x = \frac{\pi}{3} or 2π3\frac{2\pi}{3}.

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