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(D) Given the equation: $	an 	heta + 5 \cot 	heta = \sec 	heta$. Converting to $\sin 	heta$ and $\cos 	heta$: $\frac{\sin 	heta}{\cos 	heta} +

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$\phi$

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(D) Given the equation: $	an 	heta + 5 \cot 	heta = \sec 	heta$. Converting to $\sin 	heta$ and $\cos 	heta$: $\frac{\sin 	heta}{\cos 	heta} + \frac{5 \cos 	heta}{\sin 	heta} = \frac{1}{\cos 	heta}$,where $\sin 	heta 
eq 0$ and $\cos 	heta 
eq 0$. Multiplying by $\sin 	heta \cos 	heta$: $\sin^2 	heta + 5 \cos^2 	heta = \sin 	heta$. Using $\cos^2 	heta = 1 - \sin^2 	heta$: $\sin^2 	heta + 5(1 - \sin^2 	heta) = \sin 	heta$. Rearranging the terms: $4 \sin^2 	heta + \sin 	heta - 5 = 0$. Factoring the quadratic equation: $(4 \sin 	heta + 5)(\sin 	heta - 1) = 0$. This gives $\sin 	heta = -\frac{5}{4}$ or $\sin 	heta = 1$. Since $-1 \leq \sin 	heta \leq 1$,we discard $\sin 	heta = -\frac{5}{4}$. For $\sin 	heta = 1$,we have $	heta = 2n\pi + \frac{\pi}{2}$. However,the original equation requires $\cos 	heta 
eq 0$. If $\sin 	heta = 1$,then $\cos 	heta = 0$,which is undefined for $	an 	heta$ and $\sec 	heta$. Therefore,there are no solutions,and the solution set is $\phi$.

The solution set of the trigonometric equation $\tan \theta + 5 \cot \theta = \sec \theta$ is

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