If sin ^{2} theta+3 cos theta=2, then cos ^{3 | vedclass

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(D) Given, $\sin ^{2} 	heta+3 \cos 	heta=2$ $\Rightarrow 1-\cos ^{2} 	heta+3 \cos 	heta=2$ $\Rightarrow \cos ^{2} 	heta-3 \cos 	heta+1=0$ Using

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$18$

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(D) Given, $\sin ^{2} 	heta+3 \cos 	heta=2$ $\Rightarrow 1-\cos ^{2} 	heta+3 \cos 	heta=2$ $\Rightarrow \cos ^{2} 	heta-3 \cos 	heta+1=0$ Using the quadratic formula, $\cos 	heta = \frac{3 \pm \sqrt{9-4}}{2} = \frac{3 \pm \sqrt{5}}{2}$ Since $-1 \leq \cos 	heta \leq 1$ and $\frac{3+\sqrt{5}}{2} > 1$, we must have $\cos 	heta = \frac{3-\sqrt{5}}{2}$. Then $\sec 	heta = \frac{1}{\cos 	heta} = \frac{2}{3-\sqrt{5}} = \frac{2(3+\sqrt{5})}{9-5} = \frac{3+\sqrt{5}}{2}$. Now, $\cos 	heta + \sec 	heta = \frac{3-\sqrt{5}}{2} + \frac{3+\sqrt{5}}{2} = \frac{6}{2} = 3$. Also, $\cos 	heta \cdot \sec 	heta = 1$. We know that $\cos ^{3} 	heta+\sec ^{3} 	heta = (\cos 	heta+\sec 	heta)^{3} - 3 \cos 	heta \sec 	heta (\cos 	heta+\sec 	heta)$. Substituting the values: $(3)^{3} - 3(1)(3) = 27 - 9 = 18$.

If $\sin ^{2} \theta+3 \cos \theta=2,$ then $\cos ^{3} \theta+\sec ^{3} \theta$ is equal to

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