To solve the equation \(\sin(3x + 10°) = \cos(2x - 10°)\), we can use the relationship between sine and cosine of complementary angles, which states that \(\sin(A) = \cos(90° - A)\).
Using this relationship, we can re-write the cosine term:
\[ \cos(2x - 10°) = \sin(90° - (2x - 10°)) = \sin(100° - 2x) \]
So, we can rewrite the original equation as:
\[ \sin(3x + 10°) = \sin(100° - 2x) \]
Now, if \(\sin(A) = \sin(B)\), then either:
- \(A = B + k \cdot 180°\) for some integer \(k\), or
- \(A = 180° - B + k \cdot 180°\) for some integer \(k\).
Case 1: \[ 3x + 10° = 100° - 2x + k \cdot 180° \] Solving for \(x\): \[ 3x + 2x = 100° - 10° + k \cdot 180° \] \[ 5x = 90° + k \cdot 180° \] \[ x = 18° + k \cdot 36° \]
Case 2: \[ 3x + 10° = 180° - (100° - 2x) + k \cdot 180° \] This simplifies to: \[ 3x + 10° = 80° + 2x + k \cdot 180° \] Solving for \(x\): \[ 3x - 2x = 80° - 10° + k \cdot 180° \] \[ x = 70° + k \cdot 180° \]
Next, we consider the values of \(x\) obtained from \( k = 0 \):
From the first case, when \(k = 0\): \[ x = 18° \]
From the second case, when \(k = 0\): \[ x = 70° \]
Given the options \(36°, 16°, 22°, 18°\), the only possible value from our calculations is:
18°.