To solve the equation \( \sin(2x) = \cos(4x + 12^\circ) \), we can make use of the co-function identity which states that \( \sin(\theta) = \cos(90^\circ - \theta) \).
This means we can set up the equation:
\[ 2x = 90^\circ - (4x + 12^\circ) \]
Simplifying this, we get:
\[ 2x = 90^\circ - 4x - 12^\circ \]
Combine like terms:
\[ 2x + 4x = 90^\circ - 12^\circ \]
\[ 6x = 78^\circ \]
Now, divide both sides by 6:
\[ x = 13^\circ \]
Since we're looking for the acute angles in a right triangle, we now calculate \( 2x \):
\[ 2x = 2 \times 13^\circ = 26^\circ \]
Next, we can find \( 4x + 12^\circ \):
\[ 4x + 12^\circ = 4 \times 13^\circ + 12^\circ = 52^\circ + 12^\circ = 64^\circ \]
In a right triangle, the angles must sum to \( 180^\circ \) and one angle is \( 90^\circ \). Thus, the remaining two angles \( 26^\circ \) and \( 64^\circ \) are the acute angles.
The measure of the larger acute angle is:
\[ \boxed{64^\circ} \]
1 answer