If $α < β, 0 \leq α, β \leq 2 π$ and $α, β$ satisfy the equation $sin 2 x + 3 sin x = 1 + 3 cos x,$ then which of the following is/are true?

α + β = \frac{π}{2}

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α + β = \frac{3 π}{2}

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cos α + cos β = 0

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sin α + sin β = 0

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Solution

The correct options are
B $α + β = \frac{3 π}{2}$
C $cos α + cos β = 0$
D $sin α + sin β = 0$
Given equation,
$sin 2 x + 3 sin x = 1 + 3 cos x,$
$\Rightarrow sin 2 x + 3 (sin x - cos x) - 1 = 0 \dots (1)$

Let $sin x - cos x = t$
$\Rightarrow sin 2 x = 1 - t^{2}$

From $(1),$
$1 - t^{2} + 3 t - 1 = 0 \Rightarrow t^{2} - 3 t = 0$
$\Rightarrow t = 0 or t = 3$
$\Rightarrow sin x - cos x = 0 or sin x - cos x = 3 (rejected)$
$\Rightarrow sin x = cos x \Rightarrow tan x = 1 = tan \frac{π}{4}$
$\Rightarrow x = n π + \frac{π}{4}, n \in Z$
$\Rightarrow α = \frac{π}{4}, β = \frac{5 π}{4} (∵ α < β, 0 \leq α, β \leq 2 π)$

$α + β = \frac{3 π}{2}$

$cos α + cos β = cos \frac{π}{4} + cos \frac{5 π}{4} = 0$

$sin α + sin β = sin \frac{π}{4} + sin \frac{5 π}{4} = 0$

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