If $2 tan 3 A cos 3 A - tan 3 A + 1 = 2 cos 3 A$ , smallest magnitude of $A$ is

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Solution

$2 tan 3 A cos 3 A - tan 3 A + 1 = 2 cos 3 A$
$2 tan 3 A cos 3 A - tan 3 A - 2 cos 3 A + 1 = 0$
$tan 3 A (2 cos 3 A - 1) - 1 (2 cos 3 A - 1) = 0$
$(tan 3 A - 1) (2 cos 3 A - 1) = 0$
Either $tan 3 A - 1 = 0$ or $2 cos 3 A - 1 = 0$
$tan 3 A = 1$ or $cos 3 A = \frac{1}{2}$
$tan 3 A = tan 45$ or $cos 3 A = cos 60$
$3 A = 45$ or $3 A = 60$
$A = 15^{\circ}$ or $A = 20^{\circ}$
Thus, $A = 15^{\circ} o r 20^{\circ}$

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