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
B
nπ3
C
nπ+(1)nπ6
D
π12

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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