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Trigonometric equations

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The number of integral values of k belonging to $[0, 2] \cup [5, \infty]$ such that the equation $k \cos x - 3 \sin x = k + 1$ possess a solution is

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Solution

$k \cos x - 3 \sin x = k + 1 \Rightarrow$ $\frac{k}{\sqrt{k^{2} + 9}} \cos x - \frac{3}{\sqrt{k^{2} + 9}} \sin x = \frac{k + 1}{\sqrt{k^{2} + 9}}$
$\Rightarrow \cos (x + ϕ) = \frac{k + 1}{\sqrt{k^{2} + 9}}, w h e r e \tan ϕ = \frac{3}{k}$
$\Rightarrow - 1 \leq \frac{k + 1}{\sqrt{k^{2} + 9}} \leq 1 (∵ - 1 \leq \cos (x + ϕ) \leq 1)$
$\Rightarrow {(k + 1)}^{2} \leq k^{2} + 9 \Rightarrow 2 k + 1 \leq 9 \Rightarrow k \leq 4 ∴ I n [0, 2] \cup [5, \infty], k t a k e s t h e i n t e g e r v a l u e s 0, 1, 2 o n l y .$

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