To solve the equation \( \sin(3x + 10^\circ) = \cos(2x - 10^\circ) \), we can use the relationship between sine and cosine of complementary angles. Specifically, we know that:
\[ \sin(\theta) = \cos(90^\circ - \theta) \]
Therefore, we can rewrite the equation as:
\[ \sin(3x + 10^\circ) = \sin(90^\circ - (2x - 10^\circ)) \]
Simplifying the right side:
\[ \sin(3x + 10^\circ) = \sin(90^\circ - 2x + 10^\circ) = \sin(100^\circ - 2x) \]
This gives us two possible cases:
- \( 3x + 10^\circ = 100^\circ - 2x + 360^\circ n \) for any integer \( n \)
- \( 3x + 10^\circ = 180^\circ - (100^\circ - 2x) + 360^\circ n \) for any integer \( n \)
However, we'll focus on the principal values (n = 0).
For the first equation:
\[ 3x + 10^\circ = 100^\circ - 2x \]
Rearranging terms gives:
\[ 3x + 2x = 100^\circ - 10^\circ \] \[ 5x = 90^\circ \] \[ x = 18^\circ \]
Now we will check the second case:
For the second equation:
\[ 3x + 10^\circ = 180^\circ - 100^\circ + 2x \] \[ 3x + 10^\circ = 80^\circ + 2x \]
Rearranging gives:
\[ 3x - 2x = 80^\circ - 10^\circ \] \[ x = 70^\circ \]
This solution is outside the choices given. Therefore, we only consider \( x = 18^\circ \).
The correct answer is:
b. 18 degree