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  • Find the value of θ satisfying 1,1,sin3theta

Find the value of θ satisfying \left|\begin{array}{ccc} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{array}\right|=0

Answers (1)

Given:

\left|\begin{array}{ccc} 1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2 \end{array}\right|=0

Expand along Row 1

\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{12}(-1)^{1+2}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{23} \\ \mathrm{a}_{31} & \mathrm{a}_{33} \end{array}\right| \\ +\mathrm{a}_{13}(-1)^{1+3}\left|\begin{array}{cc} \mathrm{a}_{21} & \mathrm{a}_{22} \\ \mathrm{a}_{31} & \mathrm{a}_{32} \end{array}\right| \\ =\left|\begin{array}{lll} 1 \end{array}\left|\begin{array}{cc} 3 & \cos 2 \theta \\ -7 & -2 \end{array}\right|-1\left|\begin{array}{cc} -4 & \cos 2 \theta \\ 7 & -2 \end{array}\right|+\sin 3 \theta\left|\begin{array}{cc} -4 & 3 \\ 7 & -7 \end{array}\right|\right. \end{array}

 

\begin{align*} &\Rightarrow (1) \{ -6 - \{ (-7) \cos 2\theta \} \} - 1 \{ 8 - 7 \cos 2\theta \} + \sin 3\theta \{ 28 - 21 \} = 0 \\ &\Rightarrow -6 + 7 \cos 2\theta - 8 + 7 \cos 2\theta + 7 \sin 3\theta = 0 \\ &\Rightarrow 14 \cos 2\theta + 7 \sin 3\theta - 14 = 0 \\ &\Rightarrow 2 \cos 2\theta + \sin 3\theta - 2 = 0 \\ \end{align*}

We know, 

 

\begin{align*} &\cos 2\theta = 1 - 2\sin^2 \theta \\ &\sin 3\theta = 3\sin \theta - 4\sin^3 \theta \\ &\Rightarrow 2(1 - 2\sin^2 \theta) + (3\sin \theta - 4\sin^3 \theta) - 2 = 0 \\ &\Rightarrow 2 - 4\sin^2 \theta + 3\sin \theta - 4\sin^3 \theta - 2 = 0 \\ &\Rightarrow -2 + 4\sin^2 \theta - 3\sin \theta + 4\sin^3 \theta + 2 = 0 \\ &\Rightarrow \sin \theta (4\sin \theta - 3 + 4\sin^2 \theta) = 0 \\ \end{align*}

\begin{align*} &\Rightarrow \sin \theta (4\sin^2 \theta - 6\sin \theta + 2\sin \theta - 3) = 0 \\ &\Rightarrow \sin \theta [2\sin \theta (2\sin \theta - 3) + 1(2\sin \theta - 3)] = 0 \\ &\Rightarrow \sin \theta (2\sin \theta + 1)(2\sin \theta - 3) = 0 \\ &\Rightarrow \sin \theta = 0 \text{ or } 2\sin \theta + 1 = 0 \text{ or } 2\sin \theta - 3 = 0 \\ &\Rightarrow \theta = n\pi \text{ or } 2\sin \theta = -1 \text{ or } 2\sin \theta = 3 \\ \end{align*}

\begin{array}{l} \Rightarrow \mathrm{or} \sin \theta=-\frac{1}{2} \text { or } \sin \theta=\frac{3}{2} \\ \Rightarrow \theta=\mathrm{n} \pi \text { or } \theta=\mathrm{m} \pi+(-1)^{\mathrm{n}}\left(-\frac{\pi}{6}\right) ; \mathrm{m}, \mathrm{n} \in \mathrm{Z} \end{array}

But it is not possible to have \sin \theta=\frac{3}{2}

 

Posted by

infoexpert22

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