Discussion Forum

The value of $\theta$  for which the following  system of equations has a non-trivial solution is  $(4 \sin \theta)x-3y-z=0 , x-(6 \cos 2\theta )y+z=0$, 3x-12y+4z=0

Answer:

Explanation:

We have,

$(4 \sin \theta )x-3y+z=0$

$x-(6 \cos 2 \theta)y+z=0$

 3x-12y+4z=0

 system of equation has a non trival solution

$\therefore$    $\begin{bmatrix}4 \sin \theta &-3&1 \\1 & -6 \cos 2\theta&1\\3&-12&4 \end{bmatrix}=0$

 $R_{1}\rightarrow 4R_{1}-R_{3}$

$\begin{bmatrix}16 \sin \theta-3 &0&0 \\3 & -6 \cos 2\theta&1\\3&-12&4 \end{bmatrix}=0$

  =   $(16 \sin \theta -3)(-24 \cos 2 \theta+12)=0$

  =$(16 \sin \theta-3)(1-2 \cos \theta)=0$

 $\Rightarrow$       $16 \sin \theta  =1  or 2 \cos  \theta$=1

   $\theta =  \sin^{1}\frac{3}{16}$   or $\theta$ = $\frac{\pi}{6}$

 $\therefore$    $\sin^{-1} (\frac{3}{16})$