To solve the equation \( 5\cos(x) + 1.5x = 0 \), we can isolate \( \cos(x) \):

\[
5\cos(x) = -1.5x \quad \Rightarrow \quad \cos(x) = -\frac{1.5x}{5} = -0.3x
\]

This equation involves both a trigonometric function and a linear term, making it transcendental. We can analyze the behavior of the functions involved:

1. **Range of \( \cos(x) \)**: Since \( \cos(x) \) varies between -1 and 1, we need to restrict \( x \) to the values where \( -0.3x \) also lies within this range:

   \[
   -1 \leq -0.3x \leq 1
   \]

   This gives us two inequalities:

   \[
   0.3x \leq 1 \quad \Rightarrow \quad x \leq \frac{1}{0.3} \approx 3.33
   \]
   \[
   -1 \leq -0.3x \quad \Rightarrow \quad x \leq \frac{1}{0.3} \approx 3.33
   \]

   Since \( x \) must be non-negative, the relevant range is \( 0 \leq x \leq 3.33 \).

2. **Graphical Approach**: To find the solution, it's often helpful to plot the functions \( y = 5\cos(x) \) and \( y = -1.5x \) on the same graph and look for intersections.

3. **Numerical Methods**: We can also use numerical methods (like Newton's method) or graphing calculators to find approximate solutions within the defined interval.

Would you like to see a specific numerical method applied, or would you prefer a graphing approach to find the solutions? 

### Follow-up Questions:
1. Would you like to know about numerical methods to solve transcendental equations?
2. Do you want a detailed explanation of how to graph these functions?
3. Would you like to explore the behavior of the solution as \( x \) increases?
4. Are there specific values of \( x \) you want to evaluate in this context?
5. Would you like to discuss the implications of this equation in real-world scenarios?

**Tip:** Graphing the functions can often provide intuitive insights into where the solutions lie, especially for transcendental equations.

Discover how to solve the equation 5cos(x) + 1.5x = 0, involving trigonometric and linear terms. This problem explores key concepts in trigonometry and transcendental equations, suitable for Grades 11-12 students.

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