If theta_{1} and theta_{2} are respectively t | vedclass

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(D) Given the equation: $2 \cot^{2} 	heta - \frac{5}{\sin 	heta} + 4 = 0$.Since $\cot^{2} 	heta = \frac{\cos^{2} 	heta}{\sin^{2} 	heta} = \frac{1

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$\frac{\pi}{3}$

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(D) Given the equation: $2 \cot^{2} 	heta - \frac{5}{\sin 	heta} + 4 = 0$.Since $\cot^{2} 	heta = \frac{\cos^{2} 	heta}{\sin^{2} 	heta} = \frac{1 - \sin^{2} 	heta}{\sin^{2} 	heta}$,we substitute this into the equation:$2 \left( \frac{1 - \sin^{2} 	heta}{\sin^{2} 	heta} ight) - \frac{5}{\sin 	heta} + 4 = 0$.Multiplying by $\sin^{2} 	heta$ (where $\sin 	heta 
eq 0$):$2(1 - \sin^{2} 	heta) - 5 \sin 	heta + 4 \sin^{2} 	heta = 0$.$2 - 2 \sin^{2} 	heta - 5 \sin 	heta + 4 \sin^{2} 	heta = 0$.$2 \sin^{2} 	heta - 5 \sin 	heta + 2 = 0$.Factoring the quadratic equation:$(2 \sin 	heta - 1)(\sin 	heta - 2) = 0$.Thus,$\sin 	heta = \frac{1}{2}$ or $\sin 	heta = 2$. Since $\sin 	heta$ cannot be $2$,we have $\sin 	heta = \frac{1}{2}$.In the interval $(0, 2\pi) - \{\pi\}$,the solutions are $	heta = \frac{\pi}{6}$ and $	heta = \frac{5\pi}{6}$.Here,$	heta_{1} = \frac{\pi}{6}$ and $	heta_{2} = \frac{5\pi}{6}$.Now,calculate the integral:$I = \int_{\pi/6}^{5\pi/6} \cos^{2} 3	heta \, d	heta = \int_{\pi/6}^{5\pi/6} \frac{1 + \cos 6	heta}{2} \, d	heta$.$I = \frac{1}{2} \left[ 	heta + \frac{\sin 6	heta}{6} ight]_{\pi/6}^{5\pi/6}$.$I = \frac{1}{2} \left[ \left( \frac{5\pi}{6} + \frac{\sin 5\pi}{6} ight) - \left( \frac{\pi}{6} + \frac{\sin \pi}{6} ight) ight] = \frac{1}{2} \left( \frac{5\pi}{6} - \frac{\pi}{6} ight) = \frac{1}{2} \left( \frac{4\pi}{6} ight) = \frac{\pi}{3}$.

If $\theta_{1}$ and $\theta_{2}$ are respectively the smallest and the largest values of $\theta$ in $(0, 2\pi) - \{\pi\}$ which satisfy the equation $2 \cot^{2} \theta - \frac{5}{\sin \theta} + 4 = 0$,then $\int_{\theta_{1}}^{\theta_{2}} \cos^{2} 3\theta \, d\theta$ is equal to

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