Question

The solutions of the equation $\left|\begin{array}{ccc}1+{\sin }^{2}x& {\sin }^{2}x& {\sin }^{2}x\\ {\cos }^{2}x& 1+{\cos }^{2}x& {\cos }^{2}x\\ 4\sin 2x& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0,(0<x<\pi ),$ are :

• A. $\frac{\pi }{6},\frac{5\pi }{6}$
• B. $\frac{7\pi }{12},\frac{11\pi }{12}$
• C. $\frac{5\pi }{12},\frac{7\pi }{12}$
• D. $\frac{\pi }{12},\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{7\pi }{12},\frac{11\pi }{12}$

$R_1\to R_1+R_2$
$\left|\begin{array}{ccc}2& 2& 1\\ {\cos }^{2}x& 1+{\cos }^{2}x& {\cos }^{2}x\\ 4\sin 2x& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0$

$C_1\to C_1-C_2$
$\left|\begin{array}{ccc}0& 2& 1\\ -1& 1+{\cos }^{2}x& {\cos }^{2}x\\ 0& 4\sin 2x& 1+4\sin 2x\end{array}\right|=0$

$\therefore 2+8\sin 2x-4\sin 2x=0$
$\Rightarrow \sin 2x=-\frac{1}{2}$
$\Rightarrow x=\frac{7\pi }{12},\frac{11\pi }{12}$