Let theta in [0, 4pi ] satisfy the equation ( | vedclass

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(B) Given the equation $(\sin 	heta + 2) (\sin 	heta + 3) (\sin 	heta + 4) = 6$.Since $-1 \leq \sin 	heta \leq 1$,we have $1 \leq \sin 	heta + 2

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$5$

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(B) Given the equation $(\sin 	heta + 2) (\sin 	heta + 3) (\sin 	heta + 4) = 6$.Since $-1 \leq \sin 	heta \leq 1$,we have $1 \leq \sin 	heta + 2 \leq 3$,$2 \leq \sin 	heta + 3 \leq 4$,and $3 \leq \sin 	heta + 4 \leq 5$.The product of these terms is $(\sin 	heta + 2) (\sin 	heta + 3) (\sin 	heta + 4)$.If $\sin 	heta = -1$,the product is $(1)(2)(3) = 6$.If $\sin 	heta > -1$,then $\sin 	heta + 2 > 1$,$\sin 	heta + 3 > 2$,and $\sin 	heta + 4 > 3$,so the product is strictly greater than $6$.Thus,the only solution is $\sin 	heta = -1$.In the interval $[0, 4\pi]$,$\sin 	heta = -1$ at $	heta = \frac{3\pi}{2}$ and $	heta = \frac{7\pi}{2}$.The sum of these values is $\frac{3\pi}{2} + \frac{7\pi}{2} = \frac{10\pi}{2} = 5\pi$.Comparing $5\pi$ with $k\pi$,we get $k = 5$.

Let $\theta \in [0, 4\pi ]$ satisfy the equation $(\sin \theta + 2) (\sin \theta + 3) (\sin \theta + 4) = 6$. If the sum of all the values of $\theta$ is of the form $k\pi$,then the value of $k$ is:

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