If left| {begin{array}{*{20}{c}}{1 + {{sin }^ | vedclass

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(C) Given the determinant equation: $\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}	heta }&{{{\sin }^2}	heta }&{{{\sin }^2}	heta }\{{{\cos }^2}

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$-1/2$

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(C) Given the determinant equation: $\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}	heta }&{{{\sin }^2}	heta }&{{{\sin }^2}	heta }\{{{\cos }^2}	heta }&{1 + {{\cos }^2}	heta }&{{{\cos }^2}	heta }\{4\sin 4	heta }&{4\sin 4	heta }&{1 + 4\sin 4	heta }\end{array}} ight| = 0$Apply column operations: $C_1 	o C_1 - C_3$ and $C_2 	o C_2 - C_3$:$\left| {\begin{array}{*{20}{c}}1&0&{{{\sin }^2}	heta }\{ - 1}&1&{{{\cos }^2}	heta }\{ - 1}&{ - 1}&{1 + 4\sin 4	heta }\end{array}} ight| = 0$Expanding along the first row:$1(1(1 + 4\sin 4	heta) - (-1)(\cos^2 	heta)) - 0 + \sin^2 	heta((-1)(-1) - 1(-1)) = 0$$1(1 + 4\sin 4	heta + \cos^2 	heta) + \sin^2 	heta(1 + 1) = 0$$1 + 4\sin 4	heta + \cos^2 	heta + 2\sin^2 	heta = 0$Since $\cos^2 	heta + \sin^2 	heta = 1$,we have:$1 + 4\sin 4	heta + 1 + \sin^2 	heta = 0$ (Wait,let us re-evaluate the determinant expansion).Correct expansion:$1(1 + 4\sin 4	heta + \cos^2 	heta) - 0 + \sin^2 	heta(1 + 1) = 0$$1 + 4\sin 4	heta + \cos^2 	heta + 2\sin^2 	heta = 0$$2 + 4\sin 4	heta = 0$$4\sin 4	heta = -2$$\sin 4	heta = -1/2$.

If $\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$,then $\sin 4\theta$ is equal to:

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