A set of values of theta for which the system | vedclass

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(D) For a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta = 0$. $\left|\begin{array}{ccc} \sin 3 	heta & -1

Vedclass · Accepted Answer

$n \pi+(-1)^n \frac{\pi}{6}$ (here $n$ is any integer)

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(D) For a non-trivial solution,the determinant of the coefficient matrix must be zero,i.e.,$\Delta = 0$. $\left|\begin{array}{ccc} \sin 3 	heta & -1 & 1 \ \cos 2 	heta & 4 & 3 \ 2 & 7 & 7 \end{array}ight| = 0$ Expanding along the first row: $\sin 3 	heta (28 - 21) - (-1) (7 \cos 2 	heta - 6) + 1 (7 \cos 2 	heta - 8) = 0$ $7 \sin 3 	heta + 7 \cos 2 	heta - 6 + 7 \cos 2 	heta - 8 = 0$ $7 \sin 3 	heta + 14 \cos 2 	heta - 14 = 0$ Dividing by $7$: $\sin 3 	heta + 2 \cos 2 	heta - 2 = 0$ Using $\sin 3 	heta = 3 \sin 	heta - 4 \sin^3 	heta$ and $\cos 2 	heta = 1 - 2 \sin^2 	heta$: $(3 \sin 	heta - 4 \sin^3 	heta) + 2(1 - 2 \sin^2 	heta) - 2 = 0$ $3 \sin 	heta - 4 \sin^3 	heta + 2 - 4 \sin^2 	heta - 2 = 0$ $-4 \sin^3 	heta - 4 \sin^2 	heta + 3 \sin 	heta = 0$ $-\sin 	heta (4 \sin^2 	heta + 4 \sin 	heta - 3) = 0$ $-\sin 	heta (2 \sin 	heta - 1)(2 \sin 	heta + 3) = 0$ Since $2 \sin 	heta + 3 
eq 0$ for any real $	heta$,we have: $\sin 	heta = 0 \Rightarrow 	heta = n \pi$ $\sin 	heta = \frac{1}{2} \Rightarrow 	heta = n \pi + (-1)^n \frac{\pi}{6}$ Thus,the set of values is $	heta = n \pi$ or $	heta = n \pi + (-1)^n \frac{\pi}{6}$.

$A$ set of values of $\theta$ for which the system of equations $(\sin 3 \theta) x-y+z=0$,$(\cos 2 \theta) x+4 y+3 z=0, 2 x+7 y+7 z=0$ has non-trivial solutions,is

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