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solve all completely for (b)(ii) and give all working and final answer

Let's focus on solving part **(b)(ii)** of the problem, which requires solving the trigonometric equation:

\[
6\cos(x) = 3 + 4\sin(x)
\]
for \( -360^\circ \leq x \leq 360^\circ \).

### Step-by-Step Solution:

#### Step 1: Rearrange the Equation
We want to manipulate this equation to work with one trigonometric function. Let's move all terms to one side:

\[
6\cos(x) - 4\sin(x) = 3
\]

#### Step 2: Express the Equation in a Single Trigonometric Function
We will aim to express the equation in a form using a single trigonometric function. We can use the identity:

\[
R\cos(x - \alpha) = R\cos(x)\cos(\alpha) + R\sin(x)\sin(\alpha)
\]

where \( R = \sqrt{a^2 + b^2} \) and \( \tan(\alpha) = \frac{b}{a} \).

We rewrite the equation as:

\[
6\cos(x) - 4\sin(x) = R\cos(x - \alpha)
\]

#### Step 3: Calculate \( R \) and \( \alpha \)
- \( R = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \)
- \( \tan(\alpha) = \frac{-4}{6} = -\frac{2}{3} \)

Using the tangent inverse function:

\[
\alpha = \tan^{-1}\left(-\frac{2}{3}\right)
\]

Let's leave \( \alpha \) in terms of its exact form for now (it’s approximately \( \alpha \approx -33.69^\circ \)).

#### Step 4: Solve the Equation
Now the equation becomes:

\[
2\sqrt{13} \cos(x - \alpha) = 3
\]

Dividing both sides by \( 2\sqrt{13} \):

\[
\cos(x - \alpha) = \frac{3}{2\sqrt{13}} \approx 0.415
\]

#### Step 5: Solve for \( x - \alpha \)
Now, solve \( \cos(x - \alpha) = 0.415 \):

\[
x - \alpha = \cos^{-1}(0.415)
\]

\[
x - \alpha \approx 65.43^\circ \quad \text{or} \quad x - \alpha \approx 360^\circ - 65.43^\circ = 294.57^\circ
\]

#### Step 6: Add \( \alpha \) Back
Now, add \( \alpha \approx -33.69^\circ \) to both solutions:

1. \( x = 65.43^\circ + (-33.69^\circ) \approx 31.74^\circ \)
2. \( x = 294.57^\circ + (-33.69^\circ) \approx 260.88^\circ \)

#### Step 7: Consider Other Cycles
Since the equation has a periodicity of \( 360^\circ \), we must also consider solutions for \( x \) in other cycles within the range \( -360^\circ \leq x \leq 360^\circ \).

For the negative cycle:

- \( x = 31.74^\circ - 360^\circ \approx -328.26^\circ \)
- \( x = 260.88^\circ - 360^\circ \approx -99.12^\circ \)

#### Final Solutions:
Thus, the complete set of solutions for \( x \) is:

\[
x \approx 31.74^\circ, 260.88^\circ, -328.26^\circ, -99.12^\circ
\]

### Do you need further details on any step, or have any questions?

---

Here are 5 related questions to further your understanding:
1. What is the general form for combining sine and cosine into a single trigonometric expression?
2. How does the period of trigonometric functions affect the number of solutions?
3. How is the tangent inverse function used to find angles in a right triangle?
4. What are the steps for converting between degrees and radians?
5. How can trigonometric identities simplify more complex equations?

**Tip**: Always consider the periodicity of trigonometric functions when solving equations over a specified range.

Solve part (b)(ii) completely and provide all working and the final answer.

Learn how to solve the trigonometric equation 6cos(x) = 3 + 4sin(x) over the range -360° ≤ x ≤ 360° using the method of expressing the equation in a single trigonometric function. This solution involves key trigonometric identities and provides a step-by-step guide suitable for Pre-Calculus students.

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