A value of \theta \epsilon (0,\pi /3), for which 

\begin{vmatrix} 1+cos^{2}\theta &sin^{2}\theta &4cos6\theta \\ cos^{2}\theta& 1+sin^{2}\theta &4cos6\theta \\ cos^{2}\theta& sin^{2}\theta & 1+4cos6\theta \end{vmatrix}=0, is:

 

  • Option 1)

    \frac{\pi }{9}

     

     

     

  • Option 2)

    \frac{\pi }{18}

  • Option 3)

    \frac{7\pi }{24}

  • Option 4)

    \frac{7\pi }{36}

 

Answers (1)
V Vakul

\begin{vmatrix} 1+cos^{2}\theta &sin^{2}\theta &4cos6\theta \\ cos^{2}\theta& 1+sin^{2}\theta &4cos6\theta \\ cos^{2}\theta& sin^{2}\theta & 1+4cos6\theta \end{vmatrix}=0

R_1\rightarrow R_1-R_2

R_2\rightarrow R_2-R_3

=>\begin{vmatrix} 1 &-1 &0 \\ 0 & 1 &-1 \\ cos^{2}\theta & sin^{2}\theta &1+cos6\theta \end{vmatrix}=0

C_2\rightarrow C_2+C_1

=>\begin{vmatrix} 1 &0 &0 \\ 0 & 1 &-1 \\ cos^{2}\theta & 1&1+4cos6\theta \end{vmatrix}=0

=>1+4cos6\theta+1=0

=>2cos6\theta=-1

=>cos6\theta=-\frac{1}{2}=cos\frac{2\pi}{3}

=>6\theta=2n\pi\pm \frac{2\pi}{3}

=>\theta=\frac{n\pi}{3}\pm \frac{\pi}{9}

=>\theta=\frac{\pi}{9}, \frac{2\pi}{9}


Option 1)

\frac{\pi }{9}

 

 

 

Option 2)

\frac{\pi }{18}

Option 3)

\frac{7\pi }{24}

Option 4)

\frac{7\pi }{36}

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