To solve the equation \( \sin(3x + 10^\circ) = \cos(2x - 10^\circ) \), we can utilize the complementary angle identity \( \sin(\theta) = \cos(90^\circ - \theta) \).
Using this identity, we can rewrite \( \cos(2x - 10^\circ) \) as \( \sin(90^\circ - (2x - 10^\circ)) \): \[ \cos(2x - 10^\circ) = \sin(90^\circ - 2x + 10^\circ) = \sin(100^\circ - 2x) \] So now we can set the sines equal: \[ \sin(3x + 10^\circ) = \sin(100^\circ - 2x) \]
Since the two sine functions are equal, we can set up two cases for equality:
- \( 3x + 10^\circ = 100^\circ - 2x + n \cdot 180^\circ \) for some integer \( n \)
- \( 3x + 10^\circ = 180^\circ - (100^\circ - 2x) + n \cdot 180^\circ \)
Case 1:
Starting with: \[ 3x + 10 = 100 - 2x \] Combine like terms: \[ 3x + 2x = 100 - 10 \] \[ 5x = 90 \] \[ x = 18^\circ \]
Case 2:
Now for the second case: \[ 3x + 10 = 180 - 100 + 2x \] Simplifying gives: \[ 3x + 10 = 80 + 2x \] \[ 3x - 2x = 80 - 10 \] \[ x = 70^\circ \]
Since \( x \) must be within the constraints for typical trigonometric functions (usually between 0° and 180°), we mainly consider the viable solutions which fit into the context.
After checking both possibilities, the solution is that \( x = 18^\circ \) is valid and fits criteria.
Therefore, the correct answer is: \[ \boxed{18^\circ} \]