Solve $sin 2 x + 5 cos x + 5 sin x + 1 = 0$

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Solution

$1 + sin 2 x = (sin x + cos x)^{2}$
$∴ u^{2} + 5 u = 0$
$∴ u = 0, - 5$ , $∴ u = 0 \Rightarrow tan x = - 1$
$∴ x = n π - \frac{π}{4}$
$u = - 5$ i.e., $sin x + cos x = - 5$ (not possible).

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