For what $x$ is the derivative of the function $f (x)$ equal to zero?
$f (x) = sin 3 x - \sqrt{3} cos 3 x + 3 (cos x - \sqrt{3} sin x)$

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Solution

Given $f (x) = sin 3 x - \sqrt{3} cos 3 x + 3 (cos x - \sqrt{3} sin x)$
$= 2 (\frac{1}{2} sin 3 x - \frac{\sqrt{3}}{2} cos 3 x) + 6 (\frac{1}{2} cos x - \frac{\sqrt{3}}{2} sin x)$
$2 sin (3 x - \frac{π}{3}) + 6 sin (\frac{π}{6} - x)$
$f^{'} (x) = 0$
$6 cos (3 x - \frac{π}{3}) - 6 cos (\frac{π}{6} - x) = 0$
$∴ cos (3 x - \frac{π}{3}) = cos (\frac{π}{6} - x)$
$3 x - \frac{π}{3} = 2 n π \pm (\frac{π}{6} - x)$
$(∵$ if $cos θ = cos α, θ = 2 n π \pm α)$
$(i) 3 x - \frac{π}{3} = 2 n_{1} π + \frac{π}{6} - x$
$4 x = 2 n_{1} π + \frac{π}{2}$
$∴ x = (\frac{4 n_{1} + 1}{8}) π$
$(i i) 3 x - \frac{π}{3} = 2 n_{2} π \pm (\frac{π}{6} - x)$
$2 x = 2 n_{2} π + \frac{π}{6}$
$x = \frac{(12 n_{2} + 1)}{12} π$ where $n_{1}, n_{2} \in Z$
$∴ x = (\frac{4 n_{1} + 1}{8}) π, (\frac{12 n_{2} + 1}{12}) π$

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