Let θ ∈ [0, 4π ] satisfy equation (sin, θ + 2) (sin, θ + 3) (sin, θ + 4) = 6 . If sum of all

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Trigonometrical Equations

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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

A

$6$

B

$5$

C

$4$

D

$2$

Solution

$(\sin \theta+2)(\sin \theta+3)(\sin \theta+4)=6$

L.H.S. $\leq 6$ and $R . H . S .=6$

Therefore, equality only holds if

${\sin \theta=-1 \text { or } \theta=3 \pi / 2,7 \pi / 2} $

$\therefore $ ${\rm{ Sum }} = 5\pi \Rightarrow {\rm{k}} = 5$

Standard 11

Mathematics

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