If D = left|begin{array}{ccc}1 & -cos theta & | vedclass

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(C) First,we expand the determinant $D$ along the first row:$D = 1(1 - (-\cos^2 	heta)) - (-\cos 	heta)(\cos 	heta - (- \cos 	heta)) - 1(\cos^2

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$14$

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(C) First,we expand the determinant $D$ along the first row:$D = 1(1 - (-\cos^2 	heta)) - (-\cos 	heta)(\cos 	heta - (- \cos 	heta)) - 1(\cos^2 	heta - 1)$$D = 1(1 + \cos^2 	heta) + \cos 	heta(2 \cos 	heta) - (\cos^2 	heta - 1)$$D = 1 + \cos^2 	heta + 2 \cos^2 	heta - \cos^2 	heta + 1$$D = 2 + 2 \cos^2 	heta$Since the range of $\cos 	heta$ is $[-1, 1]$,the range of $\cos^2 	heta$ is $[0, 1]$.Therefore,the range of $D = 2 + 2 \cos^2 	heta$ is $[2 + 2(0), 2 + 2(1)] = [2, 4]$.Thus,the maximum value $p = 4$ and the minimum value $q = 2$.Finally,we calculate $2p + 3q = 2(4) + 3(2) = 8 + 6 = 14$.

If $D = \left|\begin{array}{ccc}1 & -\cos \theta & -1 \\ \cos \theta & 1 & -\cos \theta \\ 1 & \cos \theta & 1\end{array}\right|$,and $p$ and $q$ are the maximum and minimum values of $D$ respectively,then the value of $2p + 3q$ is . . . . . . .

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