A) \[\frac{7\pi }{24}\]
B) \[\frac{\pi }{18}\]
C) \[\frac{\pi }{9}\]
D) \[\frac{7\pi }{36}\]
Correct Answer: C
Solution :
\[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}}\] \[\left| \begin{matrix} 1 & -1 & 0 \\ {{\cos }^{2}}\theta & 1+{{\sin }^{2}}\theta & 4\cos 6\theta \\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\ \end{matrix} \right|=0\] \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\] \[\left| \begin{matrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ {{\cos }^{2}}\theta & {{\sin }^{2}}\theta & 1+4\cos 6\theta \\ \end{matrix} \right|=0\] \[\Rightarrow (1+4cos6\theta )+si{{n}^{2}}\theta +1(co{{s}^{2}}\theta )=0\] \[1+2cos6\theta =0\Rightarrow \cos 6\theta =-1/2\] \[6\theta =\frac{2\pi }{3}\Rightarrow \]
You need to login to perform this action.
You will be redirected in
3 sec
