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Trigonometric equations

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Question

If the equation $x^{2} + 5 + 4 \cos (α + β) = 2 x$ has at least one solution where $α, β \in [2, 5]$ then the value of $α + β$ is equal to

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Solution

$x^{2} + 5 + 4 \cos (α + β) = 2 x$
$\Rightarrow {(x - 1)}^{2} + 4 = - 4 \cos (a x + β)$
Here, ${(x - 1)}^{2} + 4 \geq 4, - 4 \cos (α x + β) \leq 4$ $(∵ - 1 \leq \cos (α x + β) \leq 1)$
But, the given equation has at least one solution, then ${(x - 1)}^{2} + 4 = 4 = - 4 \cos (α x + β)$
$∴ x = 1 \Rightarrow \cos (α + β) = - 1$
$\Rightarrow α + β = 3 π (∵ α + β \in [4, 10])$

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