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The number of  \mathrm{\theta \in(0,4 \pi)} for which the system of linear equations

\mathrm{3(\sin 3 \theta) x-y+z=2} \\

\mathrm{3(\cos 2 \theta) x+4 y+3 z=3} \\

\mathrm{6 x+7 y+7 z=9}

has no solution, is :

Option: 1

6


Option: 2

7


Option: 3

8


Option: 4

9


Answers (1)

best_answer

The system of equation has no solution

\mathrm{D=\left|\begin{array}{ccc} 3 \sin 3 \theta & -1 & 1 \\ 3 \cos 2 \theta & 4 & 3 \\ 6 & 7 & 7 \end{array}\right|=0 }\\

\mathrm{21 \sin 3 \theta+42 \cos 2 \theta-42=0} \\

\mathrm{\sin 3 \theta+2 \cos 2 \theta-2=0}

Number of solution is 7in \left ( 0,4\pi \right )

Hence correct option is 2

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Nehul

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