Sin(3x + 50) + cos14 = 0

To solve the equation \( \sin(3x + 50^\circ) + \cos(14^\circ) = 0 \), we can start by rewriting it as:

\[
\sin(3x + 50^\circ) = -\cos(14^\circ)
\]

Now, recall that \( \cos(14^\circ) \) is a positive value since \( 14^\circ < 90^\circ \). Therefore, \( -\cos(14^\circ) \) will be negative. We need to find the values of \( 3x + 50^\circ \) such that \( \sin(3x + 50^\circ) = -\cos(14^\circ) \).

Next, we use the fact that the sine function achieves negative values in the third and fourth quadrants. So, we can express this as:

\[
3x + 50^\circ = 180^\circ + \theta \quad \text{(third quadrant)}
\]
or
\[
3x + 50^\circ = 360^\circ - \theta \quad \text{(fourth quadrant)}
\]

Where \( \theta = \arcsin(-\cos(14^\circ)) \) (which will actually be \( -\cos(14^\circ) \) without needing to calculate \( \theta \) explicitly). We can first compute \( \cos(14^\circ) \) for reference:

Using a calculator or trigonometric tables:

\[
\cos(14^\circ) \approx 0.9703
\]

This means we want \( \sin(3x + 50^\circ) \) to equal approximately \( -0.9703 \).

Now we calculate the angles:

1. **For the Third Quadrant:**

\[
3x + 50^\circ = 180^\circ + \theta
\]
\[
3x + 50^\circ = 180^\circ + (-14^\circ) = 166^\circ
\]

Rearranging gives:
\[
3x = 166^\circ - 50^\circ = 116^\circ
\]
\[
x = \frac{116^\circ}{3} \approx 38.67^\circ
\]

2. **For the Fourth Quadrant:**

\[
3x + 50^\circ = 360^\circ - \theta
\]
\[
3x + 50^\circ = 360^\circ + 14^\circ = 374^\circ
\]

Rearranging gives:
\[
3x = 374^\circ - 50^\circ = 324^\circ
\]
\[
x = \frac{324^\circ}{3} \approx 108^\circ
\]

Thus, the general solutions are:

1. \( x \approx 38.67^\circ + \frac{360^\circ k}{3} \) for \( k \in \mathbb{Z} \)
2. \( x \approx 108^\circ + \frac{360^\circ k}{3} \) for \( k \in \mathbb{Z} \)

In conclusion, the solutions for \( x \) are of the form:

\[
x \approx 38.67^\circ + 120^\circ k \quad \text{and} \quad x \approx 108^\circ + 120^\circ k \quad \text{for } k \in \mathbb{Z}
\]