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Question

The value of ` $$\theta$$' for which the system of linear equation in $$\mathrm{x},\ \mathrm{y},\ \mathrm{z}$$ given as  $$(\sin 3\theta)\mathrm{x}-\mathrm{y}+\mathrm{z}=0$$ 
 $$(\cos2\theta)\mathrm{x}+4\mathrm{y}+3\mathrm{z}=0\  ,2\mathrm{x}+7\mathrm{y}+7\mathrm{z}=0$$ has a non trivial solution


A
nπ2
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B
nπ3
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C
nπ+(1)nπ6
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D
π12
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Solution

The correct option is C $$n\displaystyle \pi+(-1)^{n}\frac{\pi}{6}$$
Solutions are non-trivial
$$\Rightarrow \begin{vmatrix}\sin3\theta  & -1 &1 \\  \cos2\theta& 4 & 3\\  2&  7& 7\end{vmatrix}=0$$
$$\sin3\theta (7)+1(7\cos2\theta-6 )+1(7\cos2\theta -8)=0$$
$$7\cos2\theta+7\sin3\theta+7\cos2\theta-14=0$$
$${7}(2\cos2\theta+7\sin3\theta)={14} $$
$$2(1-2\sin^{2}\theta)+3\sin\theta-4\sin^{3}\theta=2$$
$$3\sin\theta-4\sin^{3}\theta-4\sin^{2}\theta=0$$
$$\sin\theta(3-4\sin^{2}\theta-4\sin\theta)=0$$
$$ \sin\theta=0\    and\ \   4\sin^{2}\theta+4\sin\theta-3=0$$
$$\sin\theta=\dfrac{-4\pm \sqrt{16+48}}{8}$$
$$=\dfrac{-4+8}{8}$$
$$\therefore \sin\theta=\dfrac{1}{2}$$

$$\therefore \theta=n\pi+(-1)^{n}\dfrac{\pi}{6} $$ or $$   \theta=n\pi$$

Mathematics

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