Find the maximum and minimum values of $6 sin x cos x + 4 cos 2 x$

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Solution

$6 sin x cos x + 4 cos 2 x = 3 sin 2 x + 4 cos 2 x$
As $- \sqrt{a^{2} + b^{2}} \leq a sin A + b cos A \leq \sqrt{a^{2} + b^{2}}$
$∴ - \sqrt{5} \leq 3 sin 2 x + 4 cos 2 x \leq \sqrt{5}$

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