We are given the trigonometric equation:

$\cot(\theta) + 3 \csc(\theta) = 5$

We will solve this over the interval $[0^\circ, 360^\circ]$.

Step 1: Express $\cot(\theta)$ and $\csc(\theta)$ in terms of $\sin(\theta)$ and $\cos(\theta)$

We know that: $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}, \quad \csc(\theta) = \frac{1}{\sin(\theta)}$

Substitute these into the equation:

$\frac{\cos(\theta)}{\sin(\theta)} + 3 \cdot \frac{1}{\sin(\theta)} = 5$

Step 2: Combine terms over a common denominator

The left-hand side has a common denominator of $\sin(\theta)$, so we can rewrite the equation as:

$\frac{\cos(\theta) + 3}{\sin(\theta)} = 5$

Step 3: Multiply both sides by $\sin(\theta)$ (assuming $\sin(\theta) \neq 0$)

This simplifies to:

$\cos(\theta) + 3 = 5 \sin(\theta)$

Step 4: Solve for $\cos(\theta)$

Rearrange the equation:

$\cos(\theta) = 5 \sin(\theta) - 3$

Step 5: Use a trigonometric identity

Square both sides to eliminate $\theta$:

$\cos^2(\theta) = (5 \sin(\theta) - 3)^2$

Since $\cos^2(\theta) + \sin^2(\theta) = 1$, substitute $\cos^2(\theta)$:

$1 - \sin^2(\theta) = (5 \sin(\theta) - 3)^2$

Step 6: Solve the quadratic equation

Expand the right-hand side:

$1 - \sin^2(\theta) = 25 \sin^2(\theta) - 30 \sin(\theta) + 9$

Simplify:

$1 - 25 \sin^2(\theta) + \sin^2(\theta) = 9 - 30 \sin(\theta)$

$1 - 25 \sin^2(\theta) = 9 - 30 \sin(\theta)$

This simplifies to a quadratic equation:

$26 \sin^2(\theta) - 30 \sin(\theta) + 8 = 0$

Now solve this quadratic equation using the quadratic formula:

$\sin(\theta) = \frac{-(-30) \pm \sqrt{(-30)^2 - 4(26)(8)}}{2(26)}$

$\sin(\theta) = \frac{30 \pm \sqrt{900 - 832}}{52}$

$\sin(\theta) = \frac{30 \pm \sqrt{68}}{52}$

$\sin(\theta) = \frac{30 \pm 8.2462}{52}$

Thus:

$\sin(\theta) = \frac{30 + 8.2462}{52} \quad \text{or} \quad \sin(\theta) = \frac{30 - 8.2462}{52}$

$\sin(\theta) \approx 0.7364 \quad \text{or} \quad \sin(\theta) \approx 0.4172$

Step 7: Solve for $\theta$

Now, solve for $\theta$ by finding the angles where $\sin(\theta)$ takes these values in the interval $[0^\circ, 360^\circ]$.

1. For $\sin(\theta) \approx 0.7364$: $\theta \approx \sin^{-1}(0.7364) \approx 47.54^\circ \quad \text{(1st quadrant)}$ The second solution in the 2nd quadrant: $\theta \approx 180^\circ - 47.54^\circ = 132.46^\circ$

2. For $\sin(\theta) \approx 0.4172$: $\theta \approx \sin^{-1}(0.4172) \approx 24.64^\circ \quad \text{(1st quadrant)}$ The second solution in the 2nd quadrant: $\theta \approx 180^\circ - 24.64^\circ = 155.36^\circ$

Final Answer

The solutions are:

$\theta \approx 47.54^\circ, 132.46^\circ, 24.64^\circ, 155.36^\circ$

Let me know if you'd like more details or have any questions!

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Related questions:

1. How do you solve trigonometric equations involving both sine and cosine?
2. What is the general form of the quadratic formula?
3. How can you use trigonometric identities to simplify equations?
4. What are the key steps for solving quadratic equations?
5. How do inverse trigonometric functions work?

Tip: Always check the domain and range of trigonometric functions when solving equations to ensure solutions are within the required interval.