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Trigonometric equations

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Question

Sum of the solutions in $[0, 8 π]$ of the equation $\tan x + \cot x + 1 = \cos (x + \frac{π}{4})$ is $k π$ then 2K =

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Solution

$\tan x + \cot x = \cos (x + \frac{π}{4}) - 1$
$L H S \leq - 2$ or $\geq 2$
$∴$ $R H S \leq - 2$
$∴ L H S = R H S = - 2$
$∴ \cos (x + \frac{π}{4}) - 1 = - 2 \Rightarrow \cos (x + \frac{π}{4}) = - 1$
$\Rightarrow x + \frac{π}{4} = 2 n π + π \Rightarrow x = 2 n π + \frac{3 π}{4}$
$\Rightarrow x = \frac{3 π}{4}, \frac{11 π}{4}, \frac{19 π}{4}, \frac{27 π}{4}$
Sum = $15 π$

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