The value of theta lying between 0 and pi / 2 | vedclass

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Question

(A) Let the given determinant be $\Delta = 0$.Applying row operations $R_1 	o R_1 - R_3$ and $R_2 	o R_2 - R_3$:$\Delta = \left| \begin{array}{ccc}

Vedclass · Accepted Answer

$\frac{7\pi}{24}$ or $\frac{11\pi}{24}$

Vedclass · Answer

(A) Let the given determinant be $\Delta = 0$.Applying row operations $R_1 	o R_1 - R_3$ and $R_2 	o R_2 - R_3$:$\Delta = \left| \begin{array}{ccc} 1 & 0 & -1 \ 0 & 1 & -1 \ \sin^2 	heta & \cos^2 	heta & 1 + 4 \sin 4 	heta \end{array} ight| = 0$Expanding along $R_1$:$1(1 + 4 \sin 4 	heta + \cos^2 	heta) - 0 + (-1)(0 - \sin^2 	heta) = 0$$1 + 4 \sin 4 	heta + \cos^2 	heta + \sin^2 	heta = 0$Since $\sin^2 	heta + \cos^2 	heta = 1$,we get:$1 + 4 \sin 4 	heta + 1 = 0$$2 + 4 \sin 4 	heta = 0$$4 \sin 4 	heta = -2$$\sin 4 	heta = -\frac{1}{2}$Given $0 The values of $4 	heta$ for which $\sin 4 	heta = -\frac{1}{2}$ in the interval $(0, 2 \pi)$ are $4 	heta = \frac{7 \pi}{6}$ and $4 	heta = \frac{11 \pi}{6}$.Therefore,$	heta = \frac{7 \pi}{24}$ and $	heta = \frac{11 \pi}{24}$.

The value of $\theta$ lying between $0$ and $\pi / 2$ and satisfying the equation $\left| \begin{array}{ccc} 1 + \sin^2 \theta & \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4 \theta \end{array} \right| = 0$ is:

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