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Factorization Method Form to Remove Indeterminate Form

The possible ...

Question

$T h e p o s s i b l e v a l u e s o f θ \in (0, π) s u c h t h a t sin (θ) + sin (4 θ) + sin (7 θ) = 0 a r e :$

A

\frac{2 π}{9}, \frac{π}{4}, \frac{4 π}{9}, \frac{π}{2}, \frac{3 π}{4}, \frac{8 π}{9}

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B

\frac{π}{4}, \frac{5 π}{12}, \frac{π}{2}, \frac{2 π}{3}, \frac{3 π}{4}, \frac{8 π}{9}

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C

\frac{2 π}{9}, \frac{π}{4}, \frac{π}{2}, \frac{2 π}{3}, \frac{3 π}{4}, \frac{35 π}{36}

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D

\frac{2 π}{9}, \frac{π}{4}, \frac{π}{2}, \frac{2 π}{3}, \frac{3 π}{4}, \frac{8 π}{9}

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Solution

The correct option is A $\frac{2 π}{9}, \frac{π}{4}, \frac{4 π}{9}, \frac{π}{2}, \frac{3 π}{4}, \frac{8 π}{9}$
$sin (θ) + sin (4 θ) + sin (7 θ) = 0$
$\Rightarrow sin (4 θ - 4 θ) + sin (4 θ) + sin (4 θ + 3 θ) = 0$
$\Rightarrow 2 sin 4 θ cos 3 θ + sin 4 θ = 0$
$\Rightarrow sin 4 θ (2 cos 3 θ + 1) = 0$
$\Rightarrow sin 4 θ = 0$ or $cos 3 θ = \frac{- 1}{2}$

$0 \in (0, π) \Rightarrow 4 θ \in (0, 4 π)$
$∴ 4 θ = π, 2 π, 3 π$ where $sin 4 θ = 0$
$∴ θ = \frac{π}{4}, \frac{2 π}{4}, \frac{3 π}{4}$

$0 \in (0, π) \Rightarrow 3 θ \in (0, 3 π)$
$∴ 3 θ = \frac{2 π}{3}, \frac{4 π}{3}, \frac{8 π}{3}$ where $cos 3 θ = \frac{- 1}{2}$
$θ = \frac{2 π}{9}, \frac{4 π}{9}, \frac{8 π}{9}$

$∴$ Possible values of $θ$ are $\frac{π}{4}, \frac{π}{2}, \frac{3 π}{4}, \frac{2 π}{9}, \frac{4 π}{9}, \frac{8 π}{9}$

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