Let θ∈0,4π satisfy the equation sinθ+2sinθ+3sinθ+4=6 .If the sum of all the values of θ is of the form kπ , then the value of k is

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

A
6
B
5
C
4
D
2

Solution:

$W e h a v e (\sin θ + 2) (\sin θ + 3) (\sin θ + 4) = 6$

$S i n c e - 1 \leq \sin θ \leq 1, L H S \geq 6$

$B u t R H S = 6 \Rightarrow \sin θ = - 1$

$S i n c e θ \in [0, 4 π] \Rightarrow θ = \frac{3 π}{2}, \frac{7 π}{2}$

$∴ s u m o f v a l u e s o f θ = \frac{3 π}{2} + \frac{7 π}{2} = 5 π = kπ \Rightarrow k = 5$

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